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.