3.6: Differentiating Under the Integral Sign

In the previous section, we noted that if an integrand contains a parameter (denoted $\gamma$) which is independent of the integration variable (denoted $x$), then the definite integral can itself be regarded as a function of $\gamma$. It can then be shown that taking the derivative of the definite integral with respect to $\gamma$ is equivalent to taking the partial derivative of the integrand: $$\frac{d}{d\gamma} \, \left[\int_a^b dx\; f(x,\gamma)\right] = \int_a^b dx \; \frac{\partial f}{\partial \gamma}(x,\gamma).$$

This operation, called differentiating under the integral sign, was first used by Leibniz, one of the inventors of calculus. It can be applied as a technique for solving integrals, popularized by Richard Feynman in his book Surely You’re Joking, Mr. Feynman!.

Here is the method. Given a definite integral $I_0$:

1. Come up with a way to generalize the integrand, by introducing a parameter $\gamma$, such that the generalized integral becomes a function $I(\gamma)$ which reduces to the original integral $I_0$ for a particular parameter value, say $\gamma = \gamma_0$.

2. Differentiate under the integral sign. If you have chosen the generalization right, the resulting integral will be easier to solve, so...

3. Solve the integral to obtain $I'(\gamma)$.

4. Integrate $I'$ over $\gamma$ to obtain the desired integral $I(\gamma)$, and evaluate it at $\gamma_0$ to obtain the desired integral $I_0$.

An example is helpful for demonstrating this procedure. Consider the integral $$\int_{0}^\infty dx \; \frac{\sin(x)}{x}.$$ First, (i) we generalize the integral as follows (we’ll soon see why): $$I(\gamma) = \int_{0}^\infty dx \; \frac{\sin(x)}{x}\, e^{-\gamma x}.$$ The desired integral is $I(0)$. Next, (ii) differentiating under the integral gives $$I'(\gamma) = - \int_{0}^\infty dx \; \sin(x)\, e^{-\gamma x}.$$ Taking the partial derivative of the integrand with respect to $\gamma$ brought down a factor of $-x$, cancelling out the troublesome denominator. Now, (iii) we solve the new integral, which can be done by integrating by parts twice: $$\begin{align} I'(\gamma) &= \left[\cos(x)\,e^{-\gamma x}\right]_0^\infty + \gamma \int_{0}^\infty dx \; \cos(x)\, e^{-\gamma x} \\ &= -1 + \gamma \left[\sin(x)\,e^{-\gamma x}\right]_0^\infty + \gamma^2 \int_{0}^\infty dx \; \sin(x)\, e^{-\gamma x} \\ &= -1 - \gamma^2 I'(\gamma).\end{align}$$ Hence, $$I'(\gamma) = - \frac{1}{1+\gamma^2}.$$ Finally, (iv) we need to integrate this over $\gamma$. But we already saw how to do this particular integral in Section 3.4, and the result is $$I(\gamma) = A - \tan^{-1}(\gamma),$$ where $A$ is a constant of integration. When $\gamma \rightarrow \infty$, the integral must vanish, which implies that $A = \tan^{-1}(+\infty) = \pi/2$. Finally, we arrive at the result $$\int_{0}^\infty dx \; \frac{\sin(x)}{x} = I(0) = \frac{\pi}{2}.$$ When we discuss contour integration in Chapter 9, we will see a more straightforward way to do this integral.