Differentiation under the integral sign

Topics in Calculus

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Differentiation under the integral sign is a useful operation in calculus. Suppose that it is required to differentiate with respect to x the function

$F(x)=\int_{a(x)}^{b(x)}f(x,t)\,dt,$

where the functions $f(x,t)\,$ and $\frac{\partial}{\partial x}\,f(x,t)\,$ are both continuous in both $t\,$ and $x\,$ in some region of the $(t,x)\,$ plane, including $a(x)\leq t\leq b(x)\,$, $x_0\leq x\leq x_1\,$, and the functions $a(x)\,$ and $b(x)\,$ are both continuous and both have continuous derivatives for $x_0\leq x\leq x_1\,$. Then for   $\,x_0\leq x\leq x_1\,\,$:

$\frac{d}{dx}\,F(x) = f(x,b(x))\,b'(x) - f(x,a(x))\,a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}\, f(x,t)\; dt\,.$

This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The fundamental theorem of calculus is just a particular case of the above formula, for $a(x)=a\,$, a constant, $b(x)=x\,$ and $f(x,t)=f(t)\,$.

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

$I_t D_x = D_x I_t,\,$

where $D_x\,$ is the partial derivative with respect to $x\,$ and $I_t\,$ 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)

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:

$\frac{d}{dt} \int_{D(t)} F(\vec{\textbf x}, t) \,dV = \int_{D(t)} \frac{\partial}{\partial t} \,F(\vec{\textbf x}, t)\,dV + \int_{\partial D(t)} \,F(\vec{\textbf x}, t)\, \vec{\textbf v}_b \cdot \vec{\textbf n} \,dA,\,$

where $F(\vec{\textbf x}, t)\,$ is a scalar function, $D(t)\,$ and $\partial D(t)\,$ denote a time-varying connected region of $\mathbb{R}^3\,$ and its boundary, respectively, $\vec{\textbf v}_b\,$ is the Eulerian velocity of the boundary (see Lagrangian and Eulerian coordinates) and $\vec{\textbf n}\,$ is unit outwards normal.

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 $p\,$-dimensions is:^[1]

$\frac{d}{dt}\int_{D(t)}\omega=\int_{D(t)} i_{\vec{\textbf v}}(d_x\omega)+\int_{\partial D(t)} i_{\vec{\textbf v}} \omega+\int_{D(t)}\dot{\omega},\,$

where $D(t)\,$ is a time-varying domain of integration, $\omega\,$ is a $p\,$-form, $\vec{\textbf v}\,$ is the vector field of the velocity, $\vec{\textbf v}=\frac{\partial\vec{\textbf x}}{\partial t}\,$, $i\,$ denotes the interior product, $d_x\omega\,$ is the exterior derivative of $\omega\,$ with respect to the space variables only and $\dot{\omega}\,$ is the time-derivative of $\omega.\,$

Derivation of the principle of differentiation under the integral sign

A definite integral is a function of its upper limit $b\,$ and its lower limit $a.\,$

If $\int_a^b f(x)\,dx$ is a continuous function of $a\,$ or $b\,$, then, from the definition of the definite integral, $\int_a^b f(t)\;dt=F(b)-F(a),\,$

$\frac{\partial}{\partial b}\int_a^b f(x)\; dx = f(b)$   and   $\frac{\partial}{\partial a}\int_a^b f(x)\; dx = -f(a)$

since, from proof of the fundamental theorem of calculus,

$\begin{align} \frac{\partial}{\partial b}\int_a^b f(x)\; dx &= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left[ \int_a^{b+\Delta b} f(x)\,dx - \int_a^b f(x)\,dx \right] \\ &= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \int_b^{b+\Delta b} f(x)\,dx = \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left[ f(b)\, \Delta b + \mathcal{O}\left(\Delta b^2\right) \right] \\ &= f(b) \qquad \text{and} \\ \frac{\partial}{\partial a}\int_a^b f(x)\; dx &= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ \int_{a+\Delta a}^b f(x)\,dx - \int_a^b f(x)\,dx \right] \\ &= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \int_{a+\Delta a}^a f(x)\,dx = \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ -f(a)\, \Delta a + \mathcal{O}\left(\Delta a^2\right) \right] \\ &= -f(a). \end{align}$

Suppose $a\,$ and $b\,$ are constant, and that $f(x)\,$ involves a parameter $\alpha\,$ which is constant in the integration but may vary to form different integrals. Then, by the definition of a function,

$\int_a^b f(x,\alpha)\;dx=\phi(\alpha).\,$

In general, this may be differentiated by differentiating under the integral sign; i.e.,

$\frac{d\phi}{d\alpha}=\int_a^b\frac{\partial}{\partial\alpha}\,f(x,\alpha)\,dx.\,$

To prove this and, at the same time, to determine conditions under which the formula is true, we proceed as follows:

From $\int_a^b f(x,\alpha)\;dx=\phi(\alpha),\,$

$\Delta\phi=\phi(\alpha+\Delta \alpha)-\phi(\alpha)=\int_a^b f(x,\alpha+\Delta\alpha)\;dx - \int_a^b f(x,\alpha)\; dx.\,$

From the fact that $\int_a^b f_1(x)\;dx+\int_a^b f_2(x)\;dx=\int_a^b[f_1(x)+f_2(x)]\;dx,$ we have

$\Delta\phi=\int_a^b f(x,\alpha+\Delta\alpha)\;dx-\int_a^b f(x,\alpha)\;dx=\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;dx.\,$

If $f(x,\alpha)\,$ is a continuous function of $x\,$ and $\alpha_0\ \leq \alpha \leq \alpha_1\,$ when $a\le x\le b\,$ then for any $\epsilon > 0\,$ there exists $\Delta \alpha\,$ such that

$|f(x,\alpha+\Delta \alpha)-f(x,\alpha)|<\epsilon\,$ for all values of $x\in[a,b].$

(This follows from the Heine–Cantor theorem that every continuous function on a compact set is uniformly continuous.)

Therefore, from $|f(x,\alpha+\Delta\alpha)-f(x,\alpha)|<\epsilon\,$ and $\Delta \phi=\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;dx,\,$

we get $|\Delta\phi|<\epsilon\,(b-a)\,$ and the fact that $\,\phi(\alpha)\,$ is, therefore, a continuous function.

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

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

Therefore,

$\begin{align} \frac{\Delta \phi}{\Delta \alpha}&=\int_a^b\frac{f(x,\alpha+\Delta\alpha)-f(x,\alpha)}{\Delta \alpha}\;dx\\ &= \int_a^b \frac{\partial\,f(x,\alpha)}{\partial \alpha}\,dx + R\\ \text{where} \quad |R| &< \int_a^b \epsilon\; dx = \epsilon\;(b-a). \end{align}$

Now, $\epsilon\rarr0\,$ as $\Delta\alpha\rarr0\,$, therefore,

$\lim_{{\Delta \alpha} \rarr 0}\frac{\Delta\phi}{\Delta \alpha}= \frac{d\phi}{d\alpha} = \int_a^b \frac{\partial}{\partial \alpha}\,f(x,\alpha)\,dx.\,$

This is the formula we set out to prove.

Now, suppose $\int_a^b f(x,\alpha)\;dx=\phi(\alpha),\,$ where $a\,$ and $b\,$ are functions of $\alpha\,$ which take increments $\Delta a\,$ and $\Delta b,\,$ respectively, when $\alpha\,$ is increased by $\Delta\alpha.\,$ Then,

$\begin{align} \Delta\phi &=\phi(\alpha+\Delta\alpha)-\phi(\alpha)=\int_{a+\Delta a}^{b+\Delta b}f(x,\alpha+\Delta\alpha)\;dx\,-\int_a^b f(x,\alpha)\;dx\, \\ &=\int_{a+\Delta a}^af(x,\alpha+\Delta\alpha)\;dx+\int_a^bf(x,\alpha+\Delta\alpha)\;dx+\int_b^{b+\Delta b}f(x,\alpha+\Delta\alpha)\;dx\,-\int_a^b f(x,\alpha)\;dx\, \\ &=-\int_a^{a+\Delta a}\,f(x,\alpha+\Delta\alpha)\;dx+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;dx+\int_b^{b+\Delta b}\,f(x,\alpha+\Delta\alpha)\;dx.\, \end{align}$

A form of the mean value theorem, $\int_a^bf(x)\;dx=(b-a)f(\xi),\,$ where $a<\xi<b,\,$ can be applied to the first and last integrals of the formula for $\Delta\phi\,$ above, resulting in

$\Delta\phi=-\Delta a\,f(\xi_1,\alpha+\Delta\alpha)+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\;dx+\Delta b\,f(\xi_2,\alpha+\Delta\alpha).\,$

Dividing by $\Delta\alpha\,$, letting $\Delta\alpha\rarr0,\,$ noticing $\xi_1\rarr a\,$ and $\xi_2\rarr b\,$ and using the result $\frac{d\phi}{d\alpha} = \int_a^b\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,dx$ yields

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

This is the general form of the Leibniz integral rule.

Examples

Example 1

$\phi(\alpha)=\int_0^1\frac{\alpha}{x^2+\alpha^2}\;dx.$

If $\alpha=0,\,$ $\phi(\alpha)=0.\,$

If $\alpha\ne0,\,$ $\phi(\alpha)=\arctan\left(\frac{1}{\alpha}\right).\,$

The function $\frac{\alpha}{x^2+\alpha^2}\,$ is not continuous at the point $(x,\alpha)=(0,0)\,$ and the function $\phi(\alpha)\,$ has a discontinuity at $\alpha=0,\,$ because $\phi(\alpha)\,$ approaches $+\frac{\pi}{2}\,$ as $\alpha\to 0^{+}\,$ and approaches $-\frac{\pi}{2}\,$ as $\alpha\to 0^{-}.\,$

If we now differentiate $\phi(\alpha)=\int_0^1\frac{\alpha}{x^2+\alpha^2}\;dx\,$ with respect to $\alpha\,$ under the integral sign, we get

$\frac{d}{d\alpha}\,\phi(\alpha)=\int_0^1\frac{x^2-\alpha^2}{(x^2+\alpha^2)^2}\;dx=-\,\frac{x}{x^2+\alpha^2}\,\bigg|_0^1=-\frac{1}{1+\alpha^2}\,$,

which is, of course, true for all values of $\alpha\,$ except $\alpha=0.\,$

Example 2

The principle of differentiating under the integral sign may sometimes be used to evaluate a definite integral.

Consider integrating $\,\phi(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos(x)+\alpha^2)\;dx\,$  (for | α | > 1).

Now,

$\begin{align} \frac{d}{d\alpha}\,\phi(\alpha)\, &=\int_0^\pi \frac{-2\cos(x)+2\alpha}{1-2\alpha \cos(x)+\alpha^2}\;dx\, \\[8pt] &=\frac{1}{\alpha}\int_0^\pi\,\left(1-\frac{(1-\alpha)^2}{1-2\alpha \cos(x)+\alpha^2}\,\right)\,dx\, \\[8pt] &=\frac{\pi}{\alpha}-\frac{2}{\alpha}\left\{\,\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\right\}\,\bigg|_0^\pi. \end{align}$

As $x\,$ varies from $0\,$ to $\pi,\,$ $\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,$ varies through positive values from $0\,$ to $\infty\,$ when $-1<\alpha<1\,$ and $\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,$ varies through negative values from $0\,$ to $-\infty\,$ when $\alpha<-1\,$ or $\alpha>1.\,$

Hence,

$\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=\frac{\pi}{2}\,$ when $-1<\alpha<1\,$

and

$\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=-\frac{\pi}{2}\,$ when $\alpha<-1\,$ or $\alpha>1.\,$

Therefore,

$\frac{d}{d\alpha}\,\phi(\alpha)\,=0\,$ when $-1<\alpha<1\,$ and

$\frac{d}{d\alpha}\,\phi(\alpha)\,=\frac{2\pi}{\alpha}\,$ when $\alpha<-1\,$ or $\alpha>1.\,$

Upon integrating both sides with respect to $\alpha\,$, we get $\phi(\alpha)=C_1\,$ when $-1<\alpha<1\,$ and $\phi(\alpha)=2\pi \ln|\alpha|+C_2\,$ when $\alpha<-1\,$ or $\alpha>1.\,$

$C_1\,$ may be determined by setting $\alpha=0\,$ in

$\begin{align} \phi(\alpha) &= \int_0^\pi \ln(1-2\alpha \cos(x)+\alpha^2)\;dx. \\[6pt] \phi(0) &=\int_0^\pi \ln(1)\;dx\, \\[6pt] &=\int_0^\pi 0\;dx\, \\[6pt] &=0\, \end{align}$

Thus, $C_1=0.\,$ Hence, $\phi(\alpha)=0\,$ when $-1<\alpha<1.\,$

To determine $C_2\,$ in the same manner, we should need to substitute in $\phi(\alpha)=\int_0^\pi \ln(1-2\alpha \cos(x)+\alpha^2)\;dx\,$ a value of $\alpha\,$ greater numerically than 1. This is somewhat inconvenient. Instead, we substitute $\alpha=\frac{1}{\beta}\,$, where $-1<\beta<1.\,$ Then,

$\begin{align} \phi(\alpha) &=\int_0^\pi\left(\ln(1-2\beta \cos(x)+\beta^2)-2\ln|\beta|\right)\;dx\, \\[8pt] &=0-2\pi\ln|\beta|\, \\[8pt] &=2\pi\ln|\alpha|\, \end{align}$

Therefore, $C_2=0\,$ (and $\phi(\alpha)=2\pi\ln|\alpha|\,$ when $\alpha<-1\,$ or $\alpha>1.\,$)

The definition of $\phi(\alpha)\,$ is now complete:

$\phi(\alpha)=0\,$ when $-1<\alpha<1\,$ and

$\phi(\alpha)=2\pi \ln|\alpha|\,$ when $\alpha<-1\,$ or $\alpha>1.\,$

The foregoing discussion, of course, does not apply when $\alpha=\pm1,\,$ since the conditions for differentiability are not met.

Example 3

Here, we consider the integration of

$\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx,\,$

where both $a,\,b\,>\,0$, by differentiating under the integral sign.

Let us first find $\textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx.\,$

Dividing both the numerator and the denominator by $\cos^2\,x$ yields

$\begin{align} \textbf J\; &=\;\int_0^{\frac{\pi}{2}}\,\frac{\sec^2\,x}{a\,+b\,\tan^2\,x}\;dx \\[6pt] &=\,\frac{1}{b}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(\sqrt{\,\frac{a}{b}\,}\right)^2+\tan^2\,x}\;d(\tan\,x)\, \\[6pt] &=\,\frac{1}{\sqrt{\,a\,b\,}}\,\left(\tan^{-1}\left(\sqrt{\,\frac{b}{a}\,}\,\tan\,x\right)\right)\,\bigg|_0^{\frac{\pi}{2}}\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}}. \end{align}$

The limits of integration being independent of $a,\,$ $\textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx\,$ gives us

$\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\,$

whereas $\textbf J\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}}$ gives us

$\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,$

Equating these two relations then yields

$\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,$

In a similar fashion, pursuing $\frac{\partial\,\textbf J}{\partial\,b}\,$ yields

$\,\int_0^{\frac{\pi}{2}}\,\frac{\sin^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a\,b^3\,}}.\,$

Adding the two results then produces

$\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx\;=\;\frac{\pi}{4\,\sqrt{\,a\,b\,}}\left(\frac{1}{a}+\frac{1}{b}\right),\,$

which is the value of the integral $\textbf I.\,$

Note that if we define

$\textbf I_n\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^n}\;dx,\,$

it can easily be shown that

$\frac{\partial\,\textbf I_{n-1}}{\partial\,a}\,+\,\frac{\partial\,\textbf I_{n-1}}{\partial\,b}\,+\,(n-1)\cdot\textbf I_n\;=\;0.\,$

Given $\textbf I_1\,,\,$ this partial-derivative-based recursive relation (i.e., integral reduction formula) can then be utilized to compute all of the values of $\textbf I_n\,$ for $n\,>\,1$ ($\textbf I_2\,$, $\textbf I_3\,$, $\textbf I_4\,$ etc.).

Example 4

Here, we consider the integral

$\textbf I(\alpha)\;=\;\int_0^{\frac{\pi}{2}}\,\frac{\ln\,(1+\cos\alpha\,\cos\,x)}{\cos\,x}\;dx,\,$

for $\,0\,<\,\alpha\,<\,\pi.$

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

$\begin{align} \frac{d}{d\alpha}\,\textbf I(\alpha)\; &=\;\int_0^{\frac{\pi}{2}}\,\frac{\partial}{\partial\alpha}\,\left(\frac{\ln\,(1\,+\,\cos\alpha\,\cos\,x)}{\cos\,x}\right)\,dx\, \\ &=\;-\,\int_0^{\frac{\pi}{2}}\,\frac{\sin\alpha}{1+\cos\alpha\,\cos\,x}\,dx\, \\ &=\;-\,\int_0^{\frac{\pi}{2}}\,\frac{\sin\alpha}{\left(\cos^2\,\frac{x}{2}+\sin^2\,\frac{x}{2}\right)\,+\,\cos\alpha\,\left(\cos^2\,\frac{x}{2}-\sin^2\,\frac{x}{2}\right)}\,dx\, \\ &=\;-\,\frac{\sin\alpha}{1-\cos\alpha}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\cos^2\,\frac{x}{2}}\,\frac{1}{\left[\,\left(\frac{1+\cos\alpha}{1-\cos\alpha}\right)\,+\,\tan^2\,\frac{x}{2}\,\right]}\,dx\, \\ &=\;-\,\frac{2\,\sin\alpha}{1-\cos\alpha}\,\int_0^{\frac{\pi}{2}}\,\frac{\frac{1}{2}\,\sec^2\,\frac{x}{2}}{\left[\,\left(\frac{2\,\cos^2\,\frac{\alpha}{2}}{2\,\sin^2\,\frac{\alpha}{2}}\right)\,+\,\tan^2\,\frac{x}{2}\,\right]}\,dx\, \\ &=\;-\,\frac{2\left(2\,\sin\,\frac{\alpha}{2}\,\cos\,\frac{\alpha}{2}\right)}{2\,\sin^2\,\frac{\alpha}{2}}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\left[\,\left(\frac{\cos\,\frac{\alpha}{2}}{\sin\,\frac{\alpha}{2}}\right)^2\,+\,\tan^2\,\frac{x}{2}\,\right]}\,d\left(\tan\,\frac{x}{2}\right)\, \\ &=\;-\,2\,\cot\,\frac{\alpha}{2}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\left[\,\cot^2\,\frac{\alpha}{2}\,+\,\tan^2\,\frac{x}{2}\,\right]}\,d\left(\tan\,\frac{x}{2}\right)\, \\ &=\;-\,2\,\left(\tan^{-1}\,\left(\tan\,\frac{\alpha}{2}\,\tan\,\frac{x}{2}\,\right)\right)\,\bigg|_0^{\frac{\pi}{2}}\, \\ &=\;-\,\alpha\, \end{align}$

Now, when $\,\alpha\;=\;\frac{\pi}{2}\,$, we have, from $\textbf I(\alpha)\;=\;\int_0^{\frac{\pi}{2}}\,\frac{\ln\,(1+\cos\alpha\,\cos\,x)}{\cos\,x}\;dx,\,$ $\textbf I\left(\frac{\pi}{2}\right)\;=\;0.\,$

Hence,

$\begin{align} \textbf I(\alpha)\; &=\;\int_{\frac{\pi}{2}}^{\alpha}\,-\,\alpha\,d\alpha\, \\ &=\;-\,\frac{1}{2}\,\alpha^2\,\bigg|_{\frac{\pi}{2}}^{\alpha}\, \\ &=\;\frac{\pi^2}{8}\,-\,\frac{\alpha^2}{2},\, \end{align}$

which is the value of the integral $\textbf I(\alpha).\;$

Example 5

Here, we consider the integral $\,\int_0^{2\pi}\,e^{\cos\theta}\;\cos\,(\sin\theta)\;d\theta.\,$

We introduce a new variable $\,\phi,\,$ and rewrite the integral as

$\,f(\phi)\;=\;\int_0^{2\pi}\;e^{\phi\cos\theta}\;\cos(\phi\sin\theta)\;d\theta.\,$

Note that for $\,\phi\;=1,\,$ $\,f(\phi)\;=f(1)=\int_0^{2\pi}\,e^{\cos\theta}\;\cos\,(\sin\theta)\;d\theta.\,$

Thus, we proceed

$\begin{align} \frac {df}{d\phi} &= \int_0^{2\pi}\;\frac{\partial}{\partial\phi}\left(e^{\phi\cos\theta}\;\cos(\phi\sin\theta)\right)\;d\theta\, \\ &= \int_0^{2\pi}\;e^{\phi\cos\theta}\;\left(\cos\theta\cos(\phi\sin\theta)\; - \;\sin\theta\sin(\phi\sin\theta)\right)\;d\theta\, \\ &= \int_0^{2\pi}\;\frac {1}{\phi}\;\frac {\partial}{\partial\theta}\left(e^{\phi\cos\theta}\;\sin(\phi\sin\theta)\right)\;d\theta\, \\ &= \frac {1}{\phi}\;\int_0^{2\pi}\;d\left(e^{\phi\cos\theta}\;\sin(\phi\sin\theta)\right)\, \\ &= \frac {1}{\phi}\;\left(e^{\phi\cos\theta}\;\sin(\phi\sin\theta)\right)\;\bigg|_0^{2\pi}\, \\ &= 0.\, \end{align}$

From the equation for $\,f(\phi),\,$ we can see $\,f(0) = 2\pi.\,$ So, integrating both sides of $\,\frac {df}{d\phi}\,=0\,$ with respect to $\,\phi\,$ between the limits $\,0\,$ and $\,1,\,$ yields

$\,\int_{f(0)}^{f(1)}\;df = \int_{0}^1\;0\;d\phi\; = \;0\,$

$\,\Rightarrow\;f(1) - f(0) = 0\,$

$\,\Rightarrow\;f(1) - 2\pi = 0\,$

$\,\Rightarrow\;f(1) = 2\pi.\,$

which is the value of the integral $\,\int_0^{2\pi}\,e^{\cos\theta}\;\cos\,(\sin\theta)\;d\theta.\,$

Example 6

Here is an example that has variable limits. Let us try to find $\,\frac{d}{dx}\;\int_{\sin\,x}^{\cos\,x}\;\cosh\,t^2\;dt\,$. In this example, we shall simply apply the above given formula, to get

$\begin{align} \frac{d}{dx} \int_{\sin x}^{\cos x} \cosh t^2\;dt &= \cosh\left(\cos^2 x\right) \frac{d}{dx}\left(\cos x\right) - \cosh\left(\sin^2 x\right) \frac{d}{dx} \left(\sin x\right) + \int_{\sin x}^{\cos x} \frac{\partial}{\partial x}\cosh t^2\;dt \\ &= - \cosh\left(\cos^2 x\right) \sin x - \cosh\left(\sin^2 x\right) \cos x \end{align}$

where the derivative with respect to x of hyperbolic cosine t squared is 0. This is a simple example on how to use this formula for variable limits.

Other problems

There are innumerable other integrals that can be solved "quickly" using the technique of differentiation under the integral sign. For examples, to solve

$\,\int_0^\infty\;\frac{\sin\,x}{x}\;dx,\text{ pick }\int_0^\infty\;e^{-\alpha\,x}\;\frac{\sin\,x}{x}\;dx,\,$

$\,\int_0^\infty\;e^{-\left(x^2+\frac{1}{x^2}\right)}\;dx,\text{ pick }\int_0^\infty\;e^{-\left(x^2+\frac{\alpha^2}{x^2}\right)}\;dx,\,$

$\,\int_0^{\frac{\pi}{2}}\;\frac{x}{\tan\,x}\;dx,\text{ pick }\int_0^{\frac{\pi}{2}}\;\frac{\tan^{-1}(\alpha\,\tan\,x)}{\tan\,x}\;dx,\,$

$\,\int_0^{\infty}\;\frac{\ln\,(1+x^2)}{1+x^2}\;dx,\text{ pick }\int_0^{\infty}\;\frac{\ln\,(1+\alpha^2\,x^2)}{1+x^2}\;dx\,$

and for

$\,\int_0^1\;\frac{x-1}{\ln\,x}\;dx,\text{ pick }\int_0^1\;\frac{x^\alpha-1}{\ln\,x}\;dx.\,$

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"), where he mentions learning it from an old text, Advanced Calculus (1926), by Frederick S. Woods (who was a professor of mathematics in the Massachusetts Institute of Technology) while in high school. The technique was not often taught when Feynman later received his formal education in calculus and, knowing it, Feynman was able to use the technique to solve some otherwise difficult integration problems upon his arrival at graduate school at Princeton University. The direct citation from Surely You're Joking, Mr. Feynman! regarding the method of differentiation under the integral sign is as follows:

“

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.

”

See also

• Leibniz integral rule
• Differentiation of integrals

References

1. ^ Flanders, Harley (June–July 1973). "Differentiation under the integral sign". American Mathematical Monthly 80 (6): 615–627. doi:Article Link on JSTOR

• "Advanced Calculus", Frederick S. Woods, Ginn and Company, 1926.
• "Advanced Calculus", David V. Widder, Dover Publications Inc., New Ed edition (Jul 1990).

External links