3.6: Differentiating Under the Integral Sign
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In the previous section, we noted that if an integrand contains a parameter (denoted γ) which is independent of the integration variable (denoted x), then the definite integral can itself be regarded as a function of γ. It can then be shown that taking the derivative of the definite integral with respect to γ is equivalent to taking the partial derivative of the integrand: ddγ[∫badxf(x,γ)]=∫badx∂f∂γ(x,γ).
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 I0:
- Come up with a way to generalize the integrand, by introducing a parameter γ, such that the generalized integral becomes a function I(γ) which reduces to the original integral I0 for a particular parameter value, say γ=γ0.
- 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′(γ).
- Integrate I′ over γ to obtain the desired integral I(γ), and evaluate it at γ0 to obtain the desired integral I0.
An example is helpful for demonstrating this procedure. Consider the integral ∫∞0dxsin(x)x. First, (i) we generalize the integral as follows (we’ll soon see why): I(γ)=∫∞0dxsin(x)xe−γx. The desired integral is I(0). Next, (ii) differentiating under the integral gives I′(γ)=−∫∞0dxsin(x)e−γx. Taking the partial derivative of the integrand with respect to γ 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: I′(γ)=[cos(x)e−γx]∞0+γ∫∞0dxcos(x)e−γx=−1+γ[sin(x)e−γx]∞0+γ2∫∞0dxsin(x)e−γx=−1−γ2I′(γ). Hence, I′(γ)=−11+γ2. Finally, (iv) we need to integrate this over γ. But we already saw how to do this particular integral in Section 3.4, and the result is I(γ)=A−tan−1(γ), where A is a constant of integration. When γ→∞, the integral must vanish, which implies that A=tan−1(+∞)=π/2. Finally, we arrive at the result ∫∞0dxsin(x)x=I(0)=π2. When we discuss contour integration in Chapter 9, we will see a more straightforward way to do this integral.