Leibniz integral rule

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In calculus, the Leibniz integral rule for differentiation under the integral sign states that for an integral of the form

$${\displaystyle {\int }_{a(x)}^{b(x)}f(x,t)dt,}$$

where ${\displaystyle -\mathrm{\infty }<a(x),b(x)<\mathrm{\infty }}$ and the integrands are functions dependent on $x,$ the derivative of this integral is expressible as

$${\displaystyle \frac{d}{dx}\left({\int }_{a(x)}^{b(x)}f(x,t)dt\right)}$$

$=f(x,b(x))\cdot \frac{d}{dx}b(x)-f(x,a(x))\cdot \frac{d}{dx}a(x)+{\int }_{a(x)}^{b(x)}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt$
where the partial derivative ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }x}}$ indicates that inside the integral, only the variation of ${\displaystyle f(x,t)}$ with $x$ is considered in taking the derivative. It is named after Gottfried Leibniz.

In the special case where the functions $a(x)$ and $b(x)$ are constants ${\displaystyle a(x)=a}$ and ${\displaystyle b(x)=b}$ with values that do not depend on $x,$ this simplifies to:

$${\displaystyle \frac{d}{dx}\left({\int }_a^{b}f(x,t)dt\right)={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt.}$$

If ${\displaystyle a(x)=a}$ is constant and ${\displaystyle b(x)=x}$, which is another common situation (for example, in the proof of Cauchy's repeated integration formula), the Leibniz integral rule becomes:

$${\displaystyle \frac{d}{dx}\left({\int }_a^{x}f(x,t)dt\right)=f(x,x)+{\int }_a^x\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt,}$$

This important result may, under certain conditions, be used to interchange the integral and partial differential operators, and is particularly useful in the differentiation of integral transforms. An example of such is the moment generating function in probability theory, a variation of the Laplace transform, which can be differentiated to generate the moments of a random variable. Whether Leibniz's integral rule applies is essentially a question about the interchange of limits.

General form: differentiation under the integral sign

Theorem — Let $f(x,t)$ be a function such that both $f(x,t)$ and its partial derivative ${\displaystyle f_x(x,t)}$ are continuous in $t$ and $x$ in some region of the ${\displaystyle xt}$-plane, including ${\displaystyle a(x)\le t\le b(x),}$ ${\displaystyle x_0\le x\le x_1.}$ Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for ${\displaystyle x_0\le x\le x_1.}$ Then, for ${\displaystyle x_0\le x\le x_1,}$

$${\displaystyle \frac{d}{dx}\left({\int }_{a(x)}^{b(x)}f(x,t)dt\right)=f(x,b(x))\cdot \frac{d}{dx}b(x)-f(x,a(x))\cdot \frac{d}{dx}a(x)+{\int }_{a(x)}^{b(x)}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt.}$$

The right hand side may also be written using Lagrange's notation as: $f(x,b(x))b^{\mathrm{\prime }}(x)-f(x,a(x))a^{\mathrm{\prime }}(x)+{\displaystyle {\int }_{a(x)}^{b(x)}{f}_x(x,t)dt.}$

Stronger versions of the theorem only require that the partial derivative exist almost everywhere, and not that it be continuous. This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The (first) fundamental theorem of calculus is just the particular case of the above formula where ${\displaystyle a(x)=a\in \mathbb{R}}$ is constant, ${\displaystyle b(x)=x,}$ and ${\displaystyle f(x,t)=f(t)}$ does not depend on $x.$

If both upper and lower limits are taken as constants, then the formula takes the shape of an operator equation:

$${\displaystyle {\mathcal{I}}_t{\mathrm{\partial }}_x={\mathrm{\partial }}_x{\mathcal{I}}_t}$$

where ${\mathrm{\partial }}_x$ is the partial derivative with respect to $x$ and ${\displaystyle {\mathcal{I}}_t}$ is the integral operator with respect to $t$ over a fixed interval. That is, it is related to the symmetry of second derivatives, but involving integrals as well as derivatives. This case is also known as the Leibniz integral rule.

The following three basic theorems on the interchange of limits are essentially equivalent:

• the interchange of a derivative and an integral (differentiation under the integral sign; i.e., Leibniz integral rule);
• the change of order of partial derivatives;
• the change of order of integration (integration under the integral sign; i.e., Fubini's theorem).

Three-dimensional, time-dependent case

See also: § Higher dimensions

Figure 1: A vector field F(r, t) defined throughout space, and a surface Σ bounded by curve ∂Σ moving with velocity v over which the field is integrated.

A Leibniz integral rule for a two dimensional surface moving in three dimensional space is

$${\displaystyle \frac{d}{dt}{\iint }_{\mathrm{\Sigma }(t)}\mathbf{F}(\mathbf{r},t)\cdot d\mathbf{A}={\iint }_{\mathrm{\Sigma }(t)}\left({\mathbf{F}}_t(\mathbf{r},t)+\left[\mathrm{\nabla }\cdot \mathbf{F}(\mathbf{r},t)\right]\mathbf{v}\right)\cdot d\mathbf{A}-{\oint }_{\mathrm{\partial }\mathrm{\Sigma }(t)}\left[\mathbf{v}\times \mathbf{F}(\mathbf{r},t)\right]\cdot d\mathbf{s},}$$

where:

• F(r, t) is a vector field at the spatial position r at time t,
• Σ is a surface bounded by the closed curve ∂Σ,
• dA is a vector element of the surface Σ,
• ds is a vector element of the curve ∂Σ,
• v is the velocity of movement of the region Σ,
• ∇⋅ is the vector divergence,
• × is the vector cross product,
• The double integrals are surface integrals over the surface Σ, and the line integral is over the bounding curve ∂Σ.

Higher dimensions

The Leibniz integral rule can be extended to multidimensional integrals. In two and three dimensions, this rule is better known from the field of fluid dynamics as the Reynolds transport theorem:

$${\displaystyle \frac{d}{dt}{\int }_{D(t)}F(\mathbf{x},t)dV={\int }_{D(t)}\frac{\mathrm{\partial }}{\mathrm{\partial }t}F(\mathbf{x},t)dV+{\int }_{\mathrm{\partial }D(t)}F(\mathbf{x},t){\mathbf{v}}_b\cdot d\mathbf{\Sigma },}$$

where ${\displaystyle F(\mathbf{x},t)}$ is a scalar function, D(t) and ∂D(t) denote a time-varying connected region of R³ and its boundary, respectively, ${\displaystyle {\mathbf{v}}_b}$ is the Eulerian velocity of the boundary (see Lagrangian and Eulerian coordinates) and dΣ = n dS is the unit normal component of the surface element.

The general statement of the Leibniz integral rule requires concepts from differential geometry, specifically differential forms, exterior derivatives, wedge products and interior products. With those tools, the Leibniz integral rule in n dimensions is

$${\displaystyle \frac{d}{dt}{\int }_{\mathrm{\Omega }(t)}\omega ={\int }_{\mathrm{\Omega }(t)}{i}_{\mathbf{v}}(d_x\omega )+{\int }_{\mathrm{\partial }\mathrm{\Omega }(t)}{i}_{\mathbf{v}}\omega +{\int }_{\mathrm{\Omega }(t)}\dot{\omega },}$$

where Ω(t) is a time-varying domain of integration, ω is a p-form, ${\displaystyle \mathbf{v}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }t}}$ is the vector field of the velocity, ${\displaystyle i_{\mathbf{v}}}$ denotes the interior product with $\mathbf{v}$, d_xω is the exterior derivative of ω with respect to the space variables only and $\dot{\omega }$ is the time derivative of ω.

However, all of these identities can be derived from a most general statement about Lie derivatives:

$${\displaystyle {\frac{d}{dt}|}_{t=0}{\int }_{{\mathrm{im}}_{{\psi }_t}(\mathrm{\Omega })}\omega ={\int }_{\mathrm{\Omega }}{\mathcal{L}}_{\mathrm{\Psi }}\omega ,}$$

Here, the ambient manifold on which the differential form $\omega$ lives includes both space and time.

• $\mathrm{\Omega }$ is the region of integration (a submanifold) at a given instant (it does not depend on $t$, since its parametrization as a submanifold defines its position in time),
• $\mathcal{L}$ is the Lie derivative,
• $\mathrm{\Psi }$ is the spacetime vector field obtained from adding the unitary vector field in the direction of time to the purely spatial vector field $\mathbf{v}$ from the previous formulas (i.e, $\mathrm{\Psi }$ is the spacetime velocity of $\mathrm{\Omega }$),
• ${\displaystyle {\psi }_t}$ is a diffeomorphism from the one-parameter group generated by the flow of $\mathrm{\Psi }$, and
• ${\displaystyle {\text{im}}_{{\psi }_t}(\mathrm{\Omega })}$ is the image of $\mathrm{\Omega }$ under such diffeomorphism.

Something remarkable about this form, is that it can account for the case when $\mathrm{\Omega }$ changes its shape and size over time, since such deformations are fully determined by $\mathrm{\Psi }$.

Measure theory statement

Let $X$ be an open subset of $\mathbf{R}$, and $\mathrm{\Omega }$ be a measure space. Suppose ${\displaystyle f:X\times \mathrm{\Omega }\to \mathbf{R}}$ satisfies the following conditions:^[5][6][2]

1. $f(x,\omega )$ is a Lebesgue-integrable function of $\omega$ for each $x\in X$.
2. For almost all $\omega \in \mathrm{\Omega }$ , the partial derivative $f_x$ exists for all $x\in X$.
3. There is an integrable function ${\displaystyle \theta :\mathrm{\Omega }\to \mathbf{R}}$ such that $|f_x(x,\omega )|\le \theta (\omega )$ for all $x\in X$ and almost every $\omega \in \mathrm{\Omega }$.

Then, for all $x\in X$,

$${\displaystyle \frac{d}{dx}{\int }_{\mathrm{\Omega }}f(x,\omega )d\omega ={\int }_{\mathrm{\Omega }}{f}_x(x,\omega )d\omega .}$$

The proof relies on the dominated convergence theorem and the mean value theorem (details below).

Proofs

Proof of basic form

We first prove the case of constant limits of integration a and b.

We use Fubini's theorem to change the order of integration. For every x and h, such that h > 0 and both x and x +h are within [x₀,x₁], we have:

$${\displaystyle {\int }_x^{x+h}{\int }_a^b{f}_x(x,t)dtdx={\int }_a^b{\int }_x^{x+h}{f}_x(x,t)dxdt={\int }_a^b\left(f(x+h,t)-f(x,t)\right)dt={\int }_a^{b}f(x+h,t)dt-{\int }_a^{b}f(x,t)dt}$$

Note that the integrals at hand are well defined since ${\displaystyle f_x(x,t)}$ is continuous at the closed rectangle ${\displaystyle [x_0,x_1]\times [a,b]}$ and thus also uniformly continuous there; thus its integrals by either dt or dx are continuous in the other variable and also integrable by it (essentially this is because for uniformly continuous functions, one may pass the limit through the integration sign, as elaborated below).

Therefore:

$${\displaystyle \frac{{\int }_a^{b}f(x+h,t)dt-{\int }_a^{b}f(x,t)dt}{h}=\frac{1}{h}{\int }_x^{x+h}{\int }_a^b{f}_x(x,t)dtdx=\frac{F(x+h)-F(x)}{h}}$$

Where we have defined:

$${\displaystyle F(u)\equiv {\int }_{x_0}^u{\int }_a^b{f}_x(x,t)dtdx}$$

(we may replace x₀ here by any other point between x₀ and x)

F is differentiable with derivative ${\int }_a^b{f}_x(x,t)dt$, so we can take the limit where h approaches zero. For the left hand side this limit is:

$${\displaystyle \frac{d}{dx}{\int }_a^{b}f(x,t)dt}$$

For the right hand side, we get:

$${\displaystyle F^{\prime }(x)={\int }_a^b{f}_x(x,t)dt}$$

And we thus prove the desired result:

$${\displaystyle \frac{d}{dx}{\int }_a^{b}f(x,t)dt={\int }_a^b{f}_x(x,t)dt}$$

Another proof using the bounded convergence theorem

If the integrals at hand are Lebesgue integrals, we may use the bounded convergence theorem (valid for these integrals, but not for Riemann integrals) in order to show that the limit can be passed through the integral sign.

Note that this proof is weaker in the sense that it only shows that f_x(x,t) is Lebesgue integrable, but not that it is Riemann integrable. In the former (stronger) proof, if f(x,t) is Riemann integrable, then so is f_x(x,t) (and thus is obviously also Lebesgue integrable).

Let

$${\displaystyle u(x)={\int }_a^{b}f(x,t)dt.}$$

(1)

By the definition of the derivative,

$${\displaystyle u^{\prime }(x)=\underset{h\to 0}{lim}\frac{u(x+h)-u(x)}{h}.}$$

(2)

Substitute equation (1) into equation (2). The difference of two integrals equals the integral of the difference, and 1/h is a constant, so

$${\displaystyle \begin{array}{rl}{u}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{{\int }_a^{b}f(x+h,t)dt-{\int }_a^{b}f(x,t)dt}{h}\\ & =\underset{h\to 0}{lim}\frac{{\int }_a^b\left(f(x+h,t)-f(x,t)\right)dt}{h}\\ & =\underset{h\to 0}{lim}{\int }_a^b\frac{f(x+h,t)-f(x,t)}{h}dt.\end{array}}$$

We now show that the limit can be passed through the integral sign.

We claim that the passage of the limit under the integral sign is valid by the bounded convergence theorem (a corollary of the dominated convergence theorem). For each δ > 0, consider the difference quotient

$${\displaystyle f_{\delta }(x,t)=\frac{f(x+\delta ,t)-f(x,t)}{\delta }.}$$

For t fixed, the mean value theorem implies there exists z in the interval [x, x + δ] such that

$${\displaystyle f_{\delta }(x,t)=f_x(z,t).}$$

Continuity of f_x(x, t) and compactness of the domain together imply that f_x(x, t) is bounded. The above application of the mean value theorem therefore gives a uniform (independent of $t$) bound on ${\displaystyle f_{\delta }(x,t)}$. The difference quotients converge pointwise to the partial derivative f_x by the assumption that the partial derivative exists.

The above argument shows that for every sequence {δ_n} → 0, the sequence ${\displaystyle \{f_{{\delta }_n}(x,t)\}}$ is uniformly bounded and converges pointwise to f_x. The bounded convergence theorem states that if a sequence of functions on a set of finite measure is uniformly bounded and converges pointwise, then passage of the limit under the integral is valid. In particular, the limit and integral may be exchanged for every sequence {δ_n} → 0. Therefore, the limit as δ → 0 may be passed through the integral sign.

Variable limits form

For a continuous real valued function g of one real variable, and real valued differentiable functions ${\displaystyle f_1}$ and ${\displaystyle f_2}$ of one real variable,

$${\displaystyle \frac{d}{dx}\left({\int }_{f_1(x)}^{f_2(x)}g(t)dt\right)=g\left(f_2(x)\right)f_2^{\prime }(x)-g\left(f_1(x)\right)f_1^{\prime }(x).}$$

This follows from the chain rule and the First Fundamental Theorem of Calculus. Define

$${\displaystyle G(x)={\int }_{f_1(x)}^{f_2(x)}g(t)dt,}$$

and

$${\displaystyle \mathrm{\Gamma }(x)={\int }_0^{x}g(t)dt.}$$

(The lower limit just has to be some number in the domain of $g$)

Then, ${\displaystyle G(x)}$ can be written as a composition: ${\displaystyle G(x)=(\mathrm{\Gamma }\circ f_2)(x)-(\mathrm{\Gamma }\circ f_1)(x)}$. The Chain Rule then implies that

$${\displaystyle G^{\prime }(x)={\mathrm{\Gamma }}^{\prime }\left(f_2(x)\right)f_2^{\prime }(x)-{\mathrm{\Gamma }}^{\prime }\left(f_1(x)\right)f_1^{\prime }(x).}$$

By the First Fundamental Theorem of Calculus, ${\displaystyle {\mathrm{\Gamma }}^{\prime }(x)=g(x)}$. Therefore, substituting this result above, we get the desired equation:

$${\displaystyle G^{\prime }(x)=g\left(f_2(x)\right)f_2^{\prime }(x)-g\left(f_1(x)\right)f_1^{\prime }(x).}$$

Note: This form can be particularly useful if the expression to be differentiated is of the form:

$${\displaystyle {\int }_{f_1(x)}^{f_2(x)}h(x)g(t)dt}$$

Because $h(x)$ does not depend on the limits of integration, it may be moved out from under the integral sign, and the above form may be used with the Product rule, i.e.,

$${\displaystyle \frac{d}{dx}\left({\int }_{f_1(x)}^{f_2(x)}h(x)g(t)dt\right)=\frac{d}{dx}\left(h(x){\int }_{f_1(x)}^{f_2(x)}g(t)dt\right)=h^{\prime }(x){\int }_{f_1(x)}^{f_2(x)}g(t)dt+h(x)\frac{d}{dx}\left({\int }_{f_1(x)}^{f_2(x)}g(t)dt\right)}$$

General form with variable limits

Set

$${\displaystyle \phi (\alpha )={\int }_a^{b}f(x,\alpha )dx,}$$

where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα. Then,

$${\displaystyle \begin{array}{rl}\mathrm{\Delta }\phi & =\phi (\alpha +\mathrm{\Delta }\alpha )-\phi (\alpha )\\ & ={\int }_{a+\mathrm{\Delta }a}^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx-{\int }_a^{b}f(x,\alpha )dx\\ & ={\int }_{a+\mathrm{\Delta }a}^{a}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_a^{b}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_b^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx-{\int }_a^{b}f(x,\alpha )dx\\ & =-{\int }_a^{a+\mathrm{\Delta }a}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+{\int }_b^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx.\end{array}}$$

A form of the mean value theorem, ${\int }_a^{b}f(x)dx=(b-a)f(\xi )$, where a < ξ < b, may be applied to the first and last integrals of the formula for Δφ above, resulting in

$${\displaystyle \mathrm{\Delta }\phi =-\mathrm{\Delta }af({\xi }_1,\alpha +\mathrm{\Delta }\alpha )+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+\mathrm{\Delta }bf({\xi }_2,\alpha +\mathrm{\Delta }\alpha ).}$$

Divide by Δα and let Δα → 0. Notice ξ₁ → a and ξ₂ → b. We may pass the limit through the integral sign:

$${\displaystyle \underset{\mathrm{\Delta }\alpha \to 0}{lim}{\int }_a^b\frac{f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )}{\mathrm{\Delta }\alpha }dx={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx,}$$

again by the bounded convergence theorem. This yields the general form of the Leibniz integral rule,

$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx+f(b,\alpha )\frac{db}{d\alpha }-f(a,\alpha )\frac{da}{d\alpha }.}$$

Alternative proof of the general form with variable limits, using the chain rule

The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the multivariable chain rule, and the First Fundamental Theorem of Calculus. Suppose $f$ is defined in a rectangle in the ${\displaystyle x-t}$ plane, for ${\displaystyle x\in [x_1,x_2]}$ and ${\displaystyle t\in [t_1,t_2]}$. Also, assume $f$ and the partial derivative $\frac{\mathrm{\partial }f}{\mathrm{\partial }x}$ are both continuous functions on this rectangle. Suppose $a,b$ are differentiable real valued functions defined on ${\displaystyle [x_1,x_2]}$, with values in ${\displaystyle [t_1,t_2]}$ (i.e. for every ${\displaystyle x\in [x_1,x_2],a(x),b(x)\in [t_1,t_2]}$). Now, set

$${\displaystyle F(x,y)={\int }_{t_1}^{y}f(x,t)dt,\text{for}x\in [x_1,x_2]\text{and}y\in [t_1,t_2]}$$

and

$${\displaystyle G(x)={\int }_{a(x)}^{b(x)}f(x,t)dt,\text{for}x\in [x_1,x_2]}$$

Then, by properties of Definite Integrals, we can write

$${\displaystyle \begin{array}{rl}G(x)& ={\int }_{t_1}^{b(x)}f(x,t)dt-{\int }_{t_1}^{a(x)}f(x,t)dt\\ & =F(x,b(x))-F(x,a(x))\end{array}}$$

Since the functions ${\displaystyle F,a,b}$ are all differentiable (see the remark at the end of the proof), by the Multivariable Chain Rule, it follows that $G$ is differentiable, and its derivative is given by the formula:

$${\displaystyle G^{\prime }(x)=\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,b(x))+\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,b(x))b^{\prime }(x)\right)-\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,a(x))+\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,a(x))a^{\prime }(x)\right)}$$

 

Now, note that for every ${\displaystyle x\in [x_1,x_2]}$, and for every ${\displaystyle y\in [t_1,t_2]}$, we have that $\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,y)={\int }_{t_1}^y\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt$, because when taking the partial derivative with respect to $x$ of $F$, we are keeping $y$ fixed in the expression ${\int }_{t_1}^{y}f(x,t)dt$; thus the basic form of Leibniz's Integral Rule with constant limits of integration applies. Next, by the First Fundamental Theorem of Calculus, we have that $\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,y)=f(x,y)$; because when taking the partial derivative with respect to $y$ of $F$, the first variable $x$ is fixed, so the fundamental theorem can indeed be applied.

Substituting these results into the equation for ${\displaystyle G^{\prime }(x)}$ above gives:

$${\displaystyle \begin{array}{rl}{G}^{\prime }(x)& =\left({\int }_{t_1}^{b(x)}\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt+f(x,b(x))b^{\prime }(x)\right)-\left({\int }_{t_1}^{a(x)}{\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}(x,t)dt+f(x,a(x))a^{\prime }(x)\right)\\ & =f(x,b(x))b^{\prime }(x)-f(x,a(x))a^{\prime }(x)+{\int }_{a(x)}^{b(x)}\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt,\end{array}}$$

as desired.

There is a technical point in the proof above which is worth noting: applying the Chain Rule to $G$ requires that $F$ already be differentiable. This is where we use our assumptions about $f$. As mentioned above, the partial derivatives of $F$ are given by the formulas $\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,y)={\int }_{t_1}^y\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt$ and $\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,y)=f(x,y)$. Since ${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}$ is continuous, its integral is also a continuous function, and since $f$ is also continuous, these two results show that both the partial derivatives of $F$ are continuous. Since continuity of partial derivatives implies differentiability of the function, $F$ is indeed differentiable.

Three-dimensional, time-dependent form

See also: § Higher dimensions

At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid ${\displaystyle \mathbf{C}(t)}$. The function ${\displaystyle \mathbf{F}(\mathbf{r},t)}$ can be written as

$${\displaystyle \mathbf{F}(\mathbf{C}(t)+\mathbf{r}-\mathbf{C}(t),t)=\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t),}$$

with $\mathbf{I}$ independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at ${\displaystyle \mathbf{C}(t)}$. For a rigidly translating surface, the limits of integration are then independent of time, so:

$${\displaystyle \frac{d}{dt}\left({\iint }_{\mathrm{\Sigma }(t)}d{\mathbf{A}}_{\mathbf{r}}\cdot \mathbf{F}(\mathbf{r},t)\right)={\iint }_{\mathrm{\Sigma }}d{\mathbf{A}}_{\mathbf{I}}\cdot \frac{d}{dt}\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t),}$$

where the limits of integration confining the integral to the region Σ no longer are time dependent so differentiation passes through the integration to act on the integrand only:

$${\displaystyle \frac{d}{dt}\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t)={\mathbf{F}}_t(\mathbf{C}(t)+\mathbf{I},t)+\mathbf{v}\cdot \mathrm{\nabla }\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t)={\mathbf{F}}_t(\mathbf{r},t)+\mathbf{v}\cdot \mathrm{\nabla }\mathbf{F}(\mathbf{r},t),}$$

with the velocity of motion of the surface defined by

$${\displaystyle \mathbf{v}=\frac{d}{dt}\mathbf{C}(t).}$$

This equation expresses the material derivative of the field, that is, the derivative with respect to a coordinate system attached to the moving surface. Having found the derivative, variables can be switched back to the original frame of reference. We notice that (see article on curl)

$${\displaystyle \mathrm{\nabla }\times \left(\mathbf{v}\times \mathbf{F}\right)=(\mathrm{\nabla }\cdot \mathbf{F}+\mathbf{F}\cdot \mathrm{\nabla })\mathbf{v}-(\mathrm{\nabla }\cdot \mathbf{v}+\mathbf{v}\cdot \mathrm{\nabla })\mathbf{F},}$$

and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over ∂Σ:

$${\displaystyle \frac{d}{dt}\left({\iint }_{\mathrm{\Sigma }(t)}\mathbf{F}(\mathbf{r},t)\cdot d\mathbf{A}\right)={\iint }_{\mathrm{\Sigma }(t)}({\mathbf{F}}_t(\mathbf{r},t)+\left(\mathbf{F}\cdot \mathrm{\nabla }\right)\mathbf{v}+\left(\mathrm{\nabla }\cdot \mathbf{F}\right)\mathbf{v}-(\mathrm{\nabla }\cdot \mathbf{v})\mathbf{F})\cdot d\mathbf{A}-{\oint }_{\mathrm{\partial }\mathrm{\Sigma }(t)}\left(\mathbf{v}\times \mathbf{F}\right)\cdot d\mathbf{s}.}$$

The sign of the line integral is based on the right-hand rule for the choice of direction of line element ds. To establish this sign, for example, suppose the field F points in the positive z-direction, and the surface Σ is a portion of the xy-plane with perimeter ∂Σ. We adopt the normal to Σ to be in the positive z-direction. Positive traversal of ∂Σ is then counterclockwise (right-hand rule with thumb along z-axis). Then the integral on the left-hand side determines a positive flux of F through Σ. Suppose Σ translates in the positive x-direction at velocity v. An element of the boundary of Σ parallel to the y-axis, say ds, sweeps out an area vt × ds in time t. If we integrate around the boundary ∂Σ in a counterclockwise sense, vt × ds points in the negative z-direction on the left side of ∂Σ (where ds points downward), and in the positive z-direction on the right side of ∂Σ (where ds points upward), which makes sense because Σ is moving to the right, adding area on the right and losing it on the left. On that basis, the flux of F is increasing on the right of ∂Σ and decreasing on the left. However, the dot product v × F ⋅ ds = −F × v ⋅ ds = −F ⋅ v × ds. Consequently, the sign of the line integral is taken as negative.

If v is a constant,

$${\displaystyle \frac{d}{dt}{\iint }_{\mathrm{\Sigma }(t)}\mathbf{F}(\mathbf{r},t)\cdot d\mathbf{A}={\iint }_{\mathrm{\Sigma }(t)}({\mathbf{F}}_t(\mathbf{r},t)+\left(\mathrm{\nabla }\cdot \mathbf{F}\right)\mathbf{v})\cdot d\mathbf{A}-{\oint }_{\mathrm{\partial }\mathrm{\Sigma }(t)}\left(\mathbf{v}\times \mathbf{F}\right)\cdot d\mathbf{s},}$$

which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.

Alternative derivation

Lemma. One has:

$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }b}\left({\int }_a^{b}f(x)dx\right)=f(b),\frac{\mathrm{\partial }}{\mathrm{\partial }a}\left({\int }_a^{b}f(x)dx\right)=-f(a).}$$

Proof. From the proof of the fundamental theorem of calculus,

$${\displaystyle \begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }b}\left({\int }_a^{b}f(x)dx\right)& =\underset{\mathrm{\Delta }b\to 0}{lim}\frac{1}{\mathrm{\Delta }b}\left[{\int }_a^{b+\mathrm{\Delta }b}f(x)dx-{\int }_a^{b}f(x)dx\right]\\ & =\underset{\mathrm{\Delta }b\to 0}{lim}\frac{1}{\mathrm{\Delta }b}{\int }_b^{b+\mathrm{\Delta }b}f(x)dx\\ & =\underset{\mathrm{\Delta }b\to 0}{lim}\frac{1}{\mathrm{\Delta }b}\left[f(b)\mathrm{\Delta }b+O\left(\mathrm{\Delta }{b}^2\right)\right]\\ & =f(b),\end{array}}$$

and

$${\displaystyle \begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }a}\left({\int }_a^{b}f(x)dx\right)& =\underset{\mathrm{\Delta }a\to 0}{lim}\frac{1}{\mathrm{\Delta }a}\left[{\int }_{a+\mathrm{\Delta }a}^{b}f(x)dx-{\int }_a^{b}f(x)dx\right]\\ & =\underset{\mathrm{\Delta }a\to 0}{lim}\frac{1}{\mathrm{\Delta }a}{\int }_{a+\mathrm{\Delta }a}^{a}f(x)dx\\ & =\underset{\mathrm{\Delta }a\to 0}{lim}\frac{1}{\mathrm{\Delta }a}\left[-f(a)\mathrm{\Delta }a+O\left(\mathrm{\Delta }{a}^2\right)\right]\\ & =-f(a).\end{array}}$$

Suppose a and b are constant, and that f(x) involves a parameter α which is constant in the integration but may vary to form different integrals. Assume that f(x, α) is a continuous function of x and α in the compact set {(x, α) : α₀ ≤ α ≤ α₁ and a ≤ x ≤ b}, and that the partial derivative f_α(x, α) exists and is continuous. If one defines:

$${\displaystyle \phi (\alpha )={\int }_a^{b}f(x,\alpha )dx,}$$

then $\phi$ may be differentiated with respect to α by differentiating under the integral sign, i.e.,

$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx.}$$

By the Heine–Cantor theorem it is uniformly continuous in that set. In other words, for any ε > 0 there exists Δα such that for all values of x in [a, b],

$${\displaystyle |f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )|<\epsilon .}$$

On the other hand,

$${\displaystyle \begin{array}{rl}\mathrm{\Delta }\phi & =\phi (\alpha +\mathrm{\Delta }\alpha )-\phi (\alpha )\\ & ={\int }_a^{b}f(x,\alpha +\mathrm{\Delta }\alpha )dx-{\int }_a^{b}f(x,\alpha )dx\\ & ={\int }_a^b\left(f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )\right)dx\\ & \le \epsilon (b-a).\end{array}}$$

Hence φ(α) is a continuous function.

Similarly if ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )}$ exists and is continuous, then for all ε > 0 there exists Δα such that:

$${\displaystyle \mathrm{\forall }x\in [a,b],\left|\frac{f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )}{\mathrm{\Delta }\alpha }-\frac{\mathrm{\partial }f}{\mathrm{\partial }\alpha }\right|<\epsilon .}$$

Therefore,

$${\displaystyle \frac{\mathrm{\Delta }\phi }{\mathrm{\Delta }\alpha }={\int }_a^b\frac{f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )}{\mathrm{\Delta }\alpha }dx={\int }_a^b\frac{\mathrm{\partial }f(x,\alpha )}{\mathrm{\partial }\alpha }dx+R,}$$

where

$${\displaystyle |R|<{\int }_a^b\epsilon dx=\epsilon (b-a).}$$

Now, ε → 0 as Δα → 0, so

$${\displaystyle \underset{\mathrm{\Delta }\alpha \to 0}{lim}\frac{\mathrm{\Delta }\phi }{\mathrm{\Delta }\alpha }=\frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx.}$$

This is the formula we set out to prove.

Now, suppose

$${\displaystyle {\int }_a^{b}f(x,\alpha )dx=\phi (\alpha ),}$$

where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then,

$${\displaystyle \begin{array}{rl}\mathrm{\Delta }\phi & =\phi (\alpha +\mathrm{\Delta }\alpha )-\phi (\alpha )\\ & ={\int }_{a+\mathrm{\Delta }a}^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx-{\int }_a^{b}f(x,\alpha )dx\\ & ={\int }_{a+\mathrm{\Delta }a}^{a}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_a^{b}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_b^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx-{\int }_a^{b}f(x,\alpha )dx\\ & =-{\int }_a^{a+\mathrm{\Delta }a}f(x,\alpha +\mathrm{\Delta }\alpha )dx+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+{\int }_b^{b+\mathrm{\Delta }b}f(x,\alpha +\mathrm{\Delta }\alpha )dx.\end{array}}$$

A form of the mean value theorem, ${\int }_a^{b}f(x)dx=(b-a)f(\xi ),$ where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in

$${\displaystyle \mathrm{\Delta }\phi =-\mathrm{\Delta }af({\xi }_1,\alpha +\mathrm{\Delta }\alpha )+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+\mathrm{\Delta }bf({\xi }_2,\alpha +\mathrm{\Delta }\alpha ).}$$

Dividing by Δα, letting Δα → 0, noticing ξ₁ → a and ξ₂ → b and using the above derivation for

$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx}$$

yields

$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx+f(b,\alpha )\frac{\mathrm{\partial }b}{\mathrm{\partial }\alpha }-f(a,\alpha )\frac{\mathrm{\partial }a}{\mathrm{\partial }\alpha }.}$$

This is the general form of the Leibniz integral rule.

Examples

Example 1: Fixed limits

Consider the function

$${\displaystyle \phi (\alpha )={\int }_0^1\frac{\alpha }{x^2+{\alpha }^2}dx.}$$

The function under the integral sign is not continuous at the point (x, α) = (0, 0), and the function φ(α) has a discontinuity at α = 0 because φ(α) approaches ±π/2 as α → 0^±.

If we differentiate φ(α) with respect to α under the integral sign, we get

$${\displaystyle \frac{d}{d\alpha }\phi (\alpha )={\int }_0^1\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }\left(\frac{\alpha }{x^2+{\alpha }^2}\right)dx={\int }_0^1\frac{x^2-{\alpha }^2}{(x^2+{\alpha }^2{)}^2}dx={-\frac{x}{x^2+{\alpha }^2}|}_0^1=-\frac{1}{1+{\alpha }^2},}$$

which is, of course, true for all values of α except α = 0. This may be integrated (with respect to α) to find

$${\displaystyle \phi (\alpha )=\{\begin{array}{ll}0,& \alpha =0,\\ -\arctan (\alpha )+\frac{\pi }{2},& \alpha \ne 0.\end{array}}$$

Example 2: Variable limits

An example with variable limits:

$${\displaystyle \begin{array}{rl}\frac{d}{dx}{\int }_{\sin x}^{\cos x}\cosh t^{2}dt& =\cosh \left({\cos }^{2}x\right)\frac{d}{dx}(\cos x)-\cosh \left({\sin }^{2}x\right)\frac{d}{dx}(\sin x)+{\int }_{\sin x}^{\cos x}\frac{\mathrm{\partial }}{\mathrm{\partial }x}(\cosh t^2)dt\\ & =\cosh ({\cos }^{2}x)(-\sin x)-\cosh ({\sin }^{2}x)(\cos x)+0\\ & =-\cosh ({\cos }^{2}x)\sin x-\cosh ({\sin }^{2}x)\cos x.\end{array}}$$

Applications

Evaluating definite integrals

The formula

$${\displaystyle \frac{d}{dx}\left({\int }_{a(x)}^{b(x)}f(x,t)dt\right)=f(x,b(x))\cdot \frac{d}{dx}b(x)-f(x,a(x))\cdot \frac{d}{dx}a(x)+{\int }_{a(x)}^{b(x)}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt}$$

can be of use when evaluating certain definite integrals. When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration.

Example 3

Consider

$${\displaystyle \phi (\alpha )={\int }_0^{\pi }\ln \left(1-2\alpha \cos (x)+{\alpha }^2\right)dx,|\alpha |\ne 1.}$$

Now,

$${\displaystyle \begin{array}{rl}\frac{d}{d\alpha }\phi (\alpha )& ={\int }_0^{\pi }\frac{-2\cos (x)+2\alpha }{1-2\alpha \cos (x)+{\alpha }^2}dx\\ & =\frac{1}{\alpha }{\int }_0^{\pi }\left(1-\frac{1-{\alpha }^2}{1-2\alpha \cos (x)+{\alpha }^2}\right)dx\\ & ={\frac{\pi }{\alpha }-\frac{2}{\alpha }\left\{\arctan \left(\frac{1+\alpha }{1-\alpha }\tan \left(\frac{x}{2}\right)\right)\right\}|}_0^{\pi }.\end{array}}$$

As $x$ varies from ${\displaystyle 0}$ to $\pi$, we have

$${\displaystyle \{\begin{array}{ll}\frac{1+\alpha }{1-\alpha }\tan \left(\frac{x}{2}\right)\ge 0,& |\alpha |<1,\\ \frac{1+\alpha }{1-\alpha }\tan \left(\frac{x}{2}\right)\le 0,& |\alpha |>1.\end{array}}$$

Hence,

$${\displaystyle {\arctan \left(\frac{1+\alpha }{1-\alpha }\tan \left(\frac{x}{2}\right)\right)|}_0^{\pi }=\{\begin{array}{ll}\frac{\pi }{2},& |\alpha |<1,\\ -\frac{\pi }{2},& |\alpha |>1.\end{array}}$$

Therefore,

$${\displaystyle \frac{d}{d\alpha }\phi (\alpha )=\{\begin{array}{ll}0,& |\alpha |<1,\\ \frac{2\pi }{\alpha },& |\alpha |>1.\end{array}}$$

Integrating both sides with respect to $\alpha$, we get:

$${\displaystyle \phi (\alpha )=\{\begin{array}{ll}{C}_1,& |\alpha |<1,\\ 2\pi \ln |\alpha |+C_2,& |\alpha |>1.\end{array}}$$

${\displaystyle C_1=0}$ follows from evaluating ${\displaystyle \phi (0)}$:

$${\displaystyle \phi (0)={\int }_0^{\pi }\ln (1)dx={\int }_0^{\pi }0dx=0.}$$

To determine $C_2$ in the same manner, we should need to substitute in a value of $\alpha$ greater than 1 in ${\displaystyle \phi (\alpha )}$. This is somewhat inconvenient. Instead, we substitute $\alpha =\frac{1}{\beta }$, where ${\displaystyle |\beta |<1}$. Then,

$${\displaystyle \begin{array}{rl}\phi (\alpha )& ={\int }_0^{\pi }\left(\ln \left(1-2\beta \cos (x)+{\beta }^2\right)-2\ln |\beta |\right)dx\\ & ={\int }_0^{\pi }\ln \left(1-2\beta \cos (x)+{\beta }^2\right)dx-{\int }_0^{\pi }2\ln |\beta |dx\\ & =0-2\pi \ln |\beta |\\ & =2\pi \ln |\alpha |.\end{array}}$$

Therefore, ${\displaystyle C_2=0}$

The definition of ${\displaystyle \phi (\alpha )}$ is now complete:

$${\displaystyle \phi (\alpha )=\{\begin{array}{ll}0,& |\alpha |<1,\\ 2\pi \ln |\alpha |,& |\alpha |>1.\end{array}}$$

The foregoing discussion, of course, does not apply when $\alpha =\pm 1$, since the conditions for differentiability are not met.

Example 4

$${\displaystyle \mathbf{I}={\int }_0^{\pi /2}\frac{1}{(a{\cos }^{2}x+b{\sin }^{2}x{)}^2}dx,a,b>0.}$$

First we calculate:

$${\displaystyle \begin{array}{rl}\mathbf{J}& ={\int }_0^{\pi /2}\frac{1}{a{\cos }^{2}x+b{\sin }^{2}x}dx\\ & ={\int }_0^{\pi /2}\frac{\frac{1}{{\cos }^{2}x}}{a+b\frac{{\sin }^{2}x}{{\cos }^{2}x}}dx\\ & ={\int }_0^{\pi /2}\frac{{\sec }^{2}x}{a+b{\tan }^{2}x}dx\\ & =\frac{1}{b}{\int }_0^{\pi /2}\frac{1}{{\left(\sqrt{\frac{a}{b}}\right)}^2+{\tan }^{2}x}d(\tan x)\\ & ={\frac{1}{\sqrt{ab}}\arctan \left(\sqrt{\frac{b}{a}}\tan x\right)|}_0^{\pi /2}\\ & =\frac{\pi }{2\sqrt{ab}}.\end{array}}$$

The limits of integration being independent of $a$, we have:

$${\displaystyle \frac{\mathrm{\partial }\mathbf{J}}{\mathrm{\partial }a}=-{\int }_0^{\pi /2}\frac{{\cos }^{2}x}{{\left(a{\cos }^{2}x+b{\sin }^{2}x\right)}^2}dx}$$

On the other hand:

$${\displaystyle \frac{\mathrm{\partial }\mathbf{J}}{\mathrm{\partial }a}=\frac{\mathrm{\partial }}{\mathrm{\partial }a}\left(\frac{\pi }{2\sqrt{ab}}\right)=-\frac{\pi }{4\sqrt{a^{3}b}}.}$$

Equating these two relations then yields

$${\displaystyle {\int }_0^{\pi /2}\frac{{\cos }^{2}x}{{\left(a{\cos }^{2}x+b{\sin }^{2}x\right)}^2}dx=\frac{\pi }{4\sqrt{a^{3}b}}.}$$

In a similar fashion, pursuing ${\displaystyle \frac{\mathrm{\partial }\mathbf{J}}{\mathrm{\partial }b}}$ yields

$${\displaystyle {\int }_0^{\pi /2}\frac{{\sin }^{2}x}{{\left(a{\cos }^{2}x+b{\sin }^{2}x\right)}^2}dx=\frac{\pi }{4\sqrt{a{b}^3}}.}$$

Adding the two results then produces

$${\displaystyle \mathbf{I}={\int }_0^{\pi /2}\frac{1}{{\left(a{\cos }^{2}x+b{\sin }^{2}x\right)}^2}dx=\frac{\pi }{4\sqrt{ab}}\left(\frac{1}{a}+\frac{1}{b}\right),}$$

which computes ${\displaystyle \mathbf{I}}$ as desired.

This derivation may be generalized. Note that if we define

$${\displaystyle {\mathbf{I}}_n={\int }_0^{\pi /2}\frac{1}{{\left(a{\cos }^{2}x+b{\sin }^{2}x\right)}^n}dx,}$$

it can easily be shown that

$${\displaystyle (1-n){\mathbf{I}}_n=\frac{\mathrm{\partial }{\mathbf{I}}_{n-1}}{\mathrm{\partial }a}+\frac{\mathrm{\partial }{\mathbf{I}}_{n-1}}{\mathrm{\partial }b}}$$

Given ${\mathbf{I}}_1$, this integral reduction formula can be used to compute all of the values of ${\mathbf{I}}_n$ for $n>1$. Integrals like $\mathbf{I}$ and $\mathbf{J}$ may also be handled using the Weierstrass substitution.

Example 5

Here, we consider the integral

$${\displaystyle \mathbf{I}(\alpha )={\int }_0^{\pi /2}\frac{\ln (1+\cos \alpha \cos x)}{\cos x}dx,0<\alpha <\pi .}$$

Differentiating under the integral with respect to $\alpha$, we have

$${\displaystyle \begin{array}{rl}\frac{d}{d\alpha }\mathbf{I}(\alpha )& ={\int }_0^{\pi /2}\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }\left(\frac{\ln (1+\cos \alpha \cos x)}{\cos x}\right)dx\\ & =-{\int }_0^{\pi /2}\frac{\sin \alpha }{1+\cos \alpha \cos x}dx\\ & =-{\int }_0^{\pi /2}\frac{\sin \alpha }{\left({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}\right)+\cos \alpha \left({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}\right)}dx\\ & =-\frac{\sin \alpha }{1-\cos \alpha }{\int }_0^{\pi /2}\frac{1}{{\cos }^2\frac{x}{2}}\frac{1}{\frac{1+\cos \alpha }{1-\cos \alpha }+{\tan }^2\frac{x}{2}}dx\\ & =-\frac{2\sin \alpha }{1-\cos \alpha }{\int }_0^{\pi /2}\frac{\frac{1}{2}{\sec }^2\frac{x}{2}}{\frac{2{\cos }^2\frac{\alpha }{2}}{2{\sin }^2\frac{\alpha }{2}}+{\tan }^2\frac{x}{2}}dx\\ & =-\frac{2\left(2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\right)}{2{\sin }^2\frac{\alpha }{2}}{\int }_0^{\pi /2}\frac{1}{{\cot }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}d\left(\tan \frac{x}{2}\right)\\ & =-2\cot \frac{\alpha }{2}{\int }_0^{\pi /2}\frac{1}{{\cot }^2\frac{\alpha }{2}+{\tan }^2\frac{x}{2}}d\left(\tan \frac{x}{2}\right)\\ & =-2\arctan \left(\tan \frac{\alpha }{2}\tan \frac{x}{2}\right){|}_0^{\pi /2}\\ & =-\alpha .\end{array}}$$

Therefore:

$${\displaystyle \mathbf{I}(\alpha )=C-\frac{{\alpha }^2}{2}.}$$

But $\mathbf{I}\left(\frac{\pi }{2}\right)=0$ by definition so $C=\frac{{\pi }^2}{8}$ and

$${\displaystyle \mathbf{I}(\alpha )=\frac{{\pi }^2}{8}-\frac{{\alpha }^2}{2}.}$$

Example 6

Here, we consider the integral

$${\displaystyle {\int }_0^{2\pi }{e}^{\cos \theta }\cos (\sin \theta )d\theta .}$$

We introduce a new variable φ and rewrite the integral as

$${\displaystyle f(\phi )={\int }_0^{2\pi }{e}^{\phi \cos \theta }\cos (\phi \sin \theta )d\theta .}$$

When φ = 1 this equals the original integral. However, this more general integral may be differentiated with respect to $\phi$:

$${\displaystyle \begin{array}{rl}\frac{df}{d\phi }& ={\int }_0^{2\pi }\frac{\mathrm{\partial }}{\mathrm{\partial }\phi }\left(e^{\phi \cos \theta }\cos (\phi \sin \theta )\right)d\theta \\ & ={\int }_0^{2\pi }{e}^{\phi \cos \theta }\left(\cos \theta \cos (\phi \sin \theta )-\sin \theta \sin (\phi \sin \theta )\right)d\theta .\end{array}}$$

Now, fix φ, and consider the vector field on ${\displaystyle {\mathbf{R}}^2}$ defined by ${\displaystyle \mathbf{F}(x,y)=(F_1(x,y),F_2(x,y)):=(e^{\phi x}\sin (\phi y),e^{\phi x}\cos (\phi y))}$. Further, choose the positive oriented parametrization of the unit circle ${\displaystyle S^1}$ given by ${\displaystyle \mathbf{r}:[0,2\pi )\to {\mathbf{R}}^2}$, ${\displaystyle \mathbf{r}(\theta ):=(\cos \theta ,\sin \theta )}$, so that ${\displaystyle {\mathbf{r}}^{\prime }(t)=(-\sin \theta ,\cos \theta )}$. Then the final integral above is precisely

$${\displaystyle \begin{array}{rl}& {\int }_0^{2\pi }{e}^{\phi \cos \theta }\left(\cos \theta \cos (\phi \sin \theta )-\sin \theta \sin (\phi \sin \theta )\right)d\theta \\ =& {\int }_0^{2\pi }(e^{\phi \cos \theta }\sin (\phi \sin \theta ),e^{\phi \cos \theta }\cos (\phi \sin \theta ))\cdot (-\sin \theta ,\cos \theta )d\theta \\ =& {\int }_0^{2\pi }\mathbf{F}(\mathbf{r}(\theta ))\cdot {\mathbf{r}}^{\prime }(\theta )d\theta \\ =& {\oint }_{S^1}\mathbf{F}(\mathbf{r})\cdot d\mathbf{r}={\oint }_{S^1}{F}_{1}dx+F_{2}dy,\end{array}}$$

the line integral of $\mathbf{F}$ over $S^1$. By Green's Theorem, this equals the double integral

$${\displaystyle {\iint }_D\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}dA,}$$

where $D$ is the closed unit disc. Its integrand is identically 0, so ${\displaystyle df/d\phi }$ is likewise identically zero. This implies that f(φ) is constant. The constant may be determined by evaluating $f$ at $\phi =0$:

$${\displaystyle f(0)={\int }_0^{2\pi }1d\theta =2\pi .}$$

Therefore, the original integral also equals $2\pi$.

Other problems to solve

There are innumerable other integrals that can be solved using the technique of differentiation under the integral sign. For example, in each of the following cases, the original integral may be replaced by a similar integral having a new parameter $\alpha$:

$${\displaystyle \begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx& \to {\int }_0^{\mathrm{\infty }}{e}^{-\alpha x}\frac{\sin x}{x}dx,\\ {\int }_0^{\pi /2}\frac{x}{\tan x}dx& \to {\int }_0^{\pi /2}\frac{{\tan }^{-1}(\alpha \tan x)}{\tan x}dx,\\ {\int }_0^{\mathrm{\infty }}\frac{\ln (1+x^2)}{1+x^2}dx& \to {\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\alpha }^2{x}^2)}{1+x^2}dx\\ {\int }_0^1\frac{x-1}{\ln x}dx& \to {\int }_0^1\frac{x^{\alpha }-1}{\ln x}dx.\end{array}}$$

The first integral, the Dirichlet integral, is absolutely convergent for positive α but only conditionally convergent when $\alpha =0$. Therefore, differentiation under the integral sign is easy to justify when $\alpha >0$, but proving that the resulting formula remains valid when $\alpha =0$ requires some careful work.

Infinite series

The measure-theoretic version of differentiation under the integral sign also applies to summation (finite or infinite) by interpreting summation as counting measure. An example of an application is the fact that power series are differentiable in their radius of convergence.

In popular culture

Differentiation under the integral sign is mentioned in the late physicist Richard Feynman's best-selling memoir Surely You're Joking, Mr. Feynman! in the chapter "A Different Box of Tools". He describes learning it, while in high school, from an old text, Advanced Calculus (1926), by Frederick S. Woods (who was a professor of mathematics in the Massachusetts Institute of Technology). The technique was not often taught when Feynman later received his formal education in calculus, but using this technique, Feynman was able to solve otherwise difficult integration problems upon his arrival at graduate school at Princeton University:

One thing I never did learn was contour integration. I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. One day he told me to stay after class. "Feynman," he said, "you talk too much and you make too much noise. I know why. You're bored. So I'm going to give you a book. You go up there in the back, in the corner, and study this book, and when you know everything that's in this book, you can talk again." So every physics class, I paid no attention to what was going on with Pascal's Law, or whatever they were doing. I was up in the back with this book: "Advanced Calculus", by Woods. Bader knew I had studied "Calculus for the Practical Man" a little bit, so he gave me the real works—it was for a junior or senior course in college. It had Fourier series, Bessel functions, determinants, elliptic functions—all kinds of wonderful stuff that I didn't know anything about. That book also showed how to differentiate parameters under the integral sign—it's a certain operation. It turns out that's not taught very much in the universities; they don't emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all their tools on it before giving the problem to me.

Palestinian 'Unity' To Destroy Israel

by Bassam Tawil
July 31, 2023 at 5:00 am

• Recently, the [Hamas] movement claimed responsibility for a number of terror attacks targeting Jewish soldiers and civilians in the West Bank. Hamas, in addition, continues to call on Palestinians to step up terror attacks with the declared intention of liberating all of "Palestine," from the Jordan River to the Mediterranean Sea, and replacing Israel with an Iranian-backed terror state.

• The Biden administration and the European Union did not condemn Abbas for meeting with the leader of a movement designated by the Americans, Europeans and other countries as a terrorist organization.

• The Biden Administration and the Europeans are also likely to continue providing financial aid to the Palestinians without asking Abbas about his repeated attempts to forge an alliance with the same Hamas terrorists who murder Jews and openly call for the elimination of Israel. The Biden Administration and the Europeans apparently have no problem with Abbas preferring to meet with the leader of a terrorist group, Hamas, than with the prime minister of Israel.

• Needless to say, the term "comprehensive resistance" is a euphemism for terrorism. As it has proven over the past three decades, Hamas outright rejects non-violent "resistance." The only "resistance" Hamas endorses is one that includes suicide bombings and the indiscriminate firing of rockets at Israeli towns and cities, as well as drive-by shootings and stabbings. The only "resistance" Hamas endorses is one that results in the murder of as many Jews as possible.

• One of Abbas's close associates, Azzam al-Ahmed, recently said that Abbas is hoping to form a unity government with Hamas. Such a move would mean that the US and European Union, the largest funders of the Palestinians, would also end up funding an Islamist movement which they have already classified as a terrorist organization. Once Hamas joins a Palestinian Authority-led government, it too would benefit from American and European taxpayer funds.

• At the very least, the Biden Administration and the European Union ought to call out Abbas for seeking to forge an alliance with one of the most deadly Palestinian terrorist movements, Hamas.

The Biden administration and the European Union did not condemn Palestinian Authority President Mahmoud Abbas for his meeting last week with Ismail Haniyeh, the leader of Hamas, a movement designated by the Americans, Europeans and other countries as a terrorist organization. Pictured: Abbas talks with Haniyeh on April 5, 2007 in Gaza City. (Photo by Mohamed Alostaz/PPM via Getty Images)

Ismail Haniyeh is the leader of Hamas, the Palestinian Islamist movement designated by the US and European Union as a terrorist organization because of its "armed resistance" against Israel. Haniyeh, who is currently based in Qatar, has always supported terror attacks against Israel and remains fully committed to Hamas's covenant, which states that the land of "Palestine" belongs only to Muslims and rejects any "initiatives or so-called peaceful solutions and international conferences." The Hamas covenant further states that "there is no solution for the Palestinian question except through Jihad [holy war]."

Haniyeh and his movement continue to boast of terror attacks against Israeli Jews, including the cold-blooded murder of a Jewish woman and her two daughters. Recently, the movement claimed responsibility for a number of terror attacks targeting Jewish soldiers and civilians in the West Bank. Hamas, in addition, continues to call on Palestinians to step up terror attacks with the declared intention of liberating all of "Palestine," from the Jordan River to the Mediterranean Sea, and replacing Israel with an Iranian-backed terror state.

Because of his radical views and support of terrorism against Israel, Haniyeh has become so popular among Palestinians that public opinion polls show that if presidential elections were held today, he would win easily. Because of his anti-Israel rhetoric and activities, Haniyeh and his movement have been warmly embraced by Turkey and Qatar. Haniyeh and other Hamas leaders, including Khaled Mashaal and Saleh Arouri, feel more comfortable in these countries than they do in the Hamas-controlled Gaza Strip. Qatar and Turkey are considered the prime backers of the Muslim Brotherhood organization, of which Hamas is an offshoot.

On July 26, Haniyeh was invited to Turkey for a meeting with Palestinian Authority (PA) President Mahmoud Abbas, who has close relations with the Biden Administration and the European Union. After the discussion, Abbas and Haniyeh were invited to the presidential palace in Ankara for an audience with Turkish President Recep Tayyip Erdogan.

The Biden administration and the European Union did not condemn Abbas for meeting with the leader of a movement designated by the Americans, Europeans and other countries as a terrorist organization. In fact, American and European officials are likely to continue to hold regular meetings with Abbas and senior PA officials despite the new harmonious relations between Abbas and Hamas. The Biden Administration and the Europeans are also likely to continue providing financial aid to the Palestinians without asking Abbas about his repeated attempts to forge an alliance with the same Hamas terrorists who murder Jews and openly call for the elimination of Israel. The Biden Administration and the Europeans apparently have no problem with Abbas preferring to meet with the leader of a terrorist group, Hamas, than with the prime minister of Israel.

Abbas and Haniyeh did not meet to discuss ways of reviving the stalled peace process between the Palestinians and Israel or to discuss ways of boosting the Palestinian economy and solving the problem of unemployment in the West Bank and Gaza Strip.

Instead, the two met to discuss ways of "unifying [Palestinian] national efforts" in the fight against Israel, according to senior Hamas official Hussam Badran. When Hamas refers to "efforts" against Israel, it is referring to shootings, stabbings, rockets and car-ramming attacks. As its covenant unambiguously states, Hamas believes that terrorism is the only way to achieve its goal of destroying the "Zionist entity."

Badran said that Haniyeh and the Hamas officials who met with Abbas in Turkey "affirmed during the meeting that comprehensive resistance is the most effective way to confront" Israel. Needless to say, the term "comprehensive resistance" is a euphemism for terrorism. As it has proven over the past three decades, Hamas outright rejects non-violent "resistance." The only "resistance" Hamas endorses is one that includes suicide bombings and the indiscriminate firing of rockets at Israeli towns and cities, as well as drive-by shootings and stabbings. The only "resistance" Hamas endorses is one that results in the murder of as many Jews as possible.

So, why did Abbas, who has repeatedly claimed that he supports the establishment of a Palestinian state alongside Israel, agree to meet with a terrorist leader who has dedicated his entire life to destroying Israel? Why did he agree to meet with a terrorist leader whose movement expelled him and the Palestinian Authority from the Gaza Strip in 2007?

Abbas is either afraid that Hamas will kill him and topple his regime in the West Bank, or he identifies with the terror group and supports its Jihad against Israel. If the former is true, Abbas is a weak leader who cares only about his survival. If the latter is true, which seems more likely, Abbas has exposed his real intention: the elimination of Israel. By embracing the Hamas leader, Abbas is actually endorsing the Islamist movement's charter, which does not believe in Israel's right to exist. The only solution Abbas and Hamas want is one that would lead to the obliteration of Israel.

Abbas and his senior officials in Ramallah, the de facto capital of the Palestinians, have argued over the years that their repeated overtures of friendship toward Hamas are only aimed at achieving Palestinian "national unity" and ending the split between the West Bank and Gaza Strip. One of Abbas's close associates, Azzam al-Ahmed, recently said that Abbas is hoping to form a unity government with Hamas. Such a move would mean that the US and European Union, the largest funders of the Palestinians, would also end up funding an Islamist movement which they have already classified as a terrorist organization. Once Hamas joins a Palestinian Authority-led government, it too would benefit from American and European taxpayer funds.

The meeting between Abbas and Haniyeh was not the first of its kind. Last year, the two met in Algeria, where they also reportedly talked about the need to achieve "national unity" as a way of strengthening the fight against Israel. As is the situation today, when their friend Abbas met with the leader of the Palestinian terrorist movement, the Biden Administration and the European Union remained silent.

The Biden Administration is evidently more concerned about an Israeli law passed by a democratic vote in the Israeli parliament than by Abbas's ongoing efforts to appease Palestinian terrorists, whether by inviting them to join his government or by rewarding them financially through his pay-for-slay program. Rather than reviling Israeli Prime Minister Benjamin Netanyahu and the Israeli parliament, perhaps the Biden Administration might ask the Palestinians why they do not have their own parliament. The Biden Administration does not ask because it does not want to embarrass the Palestinians by showing that they do not have democracy and cannot even hold elections. Americans and Europeans seems always to avoid holding the Palestinians accountable for any wrongdoing.

Is the Biden Administration aware that the Palestinian parliament was dissolved by Abbas in 2018 and that the Palestinians have no freedom of expression under either the Palestinian Authority in the West Bank or Hamas in the Gaza Strip? Is the Biden Administration aware that while Netanyahu and his coalition were elected by a majority of the Israelis, Abbas has just entered the 18^th year of his four-year term in office?

There is no love lost between the Biden Administration and Netanyahu, who has won six elections. This is the same Biden Administration that looks the other way as the Palestinian Authority president pursues his efforts to join forces with Hamas.

Similarly, the European Union, which never misses an opportunity to denounce Israel, seems unconcerned about Abbas meeting a terrorist leader and inviting him to join the Palestinian government. Yet, at a bare minimum, the Biden Administration and the European Union ought to urge Abbas to return to the negotiating table with Israel and cease his ongoing campaign to vilify Israel and demonize Jews. The negotiations will not achieve peace and harmony between Israel and the Palestinians, but they could ease tensions between the two sides and reduce the level of violence. At the very least, the Biden Administration and the European Union ought to call out Abbas for seeking to forge an alliance with one of the most deadly Palestinian terrorist movements, Hamas.

Bassam Tawil is a Muslim Arab based in the Middle East.