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Physics LibreTexts

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)]=badxfγ(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:

  1. 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.

  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(γ).

  4. 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)xeγx. The desired integral is I(0). Next, (ii) differentiating under the integral gives I(γ)=0dxsin(x)eγx. 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)eγx]0+γ0dxcos(x)eγx=1+γ[sin(x)eγx]0+γ20dxsin(x)eγx=1γ2I(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)=Atan1(γ), where A is a constant of integration. When γ, the integral must vanish, which implies that A=tan1(+)=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.


This page titled 3.6: Differentiating Under the Integral Sign is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform.

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