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 \(I_0\) :
-
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
\(\gamma = \gamma_0\)
.
-
Differentiate under the integral sign. If you have chosen the generalization right, the resulting integral will be easier to solve, so...
-
Solve the integral to obtain
\(I'(\gamma)\)
.
-
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.