Abstract
The starting point for Feynman’s trick of ‘differentiating under the integral sign,’ mentioned at the end of Chap. 1, is Leibniz’s formula. If we have the integral
where α is the so-called parameter of the integral (not the dummy variable of integration which is, of course, x), then we wish to calculate the derivative of I with respect to α. We do that in just the way you’d expect, from the very definition of the derivative:
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Notes
- 1.
The probability integral is most commonly evaluated in textbooks with the trick of converting it to a double integral in polar coordinates (see, for example, my books An Imaginary Tale: the story of \( \sqrt{-1} \), Princeton 2012, pp. 177–178, and Mrs. Perkins’s Electric Quilt, Princeton 2009, pp. 282–283), and the use of Leibniz’s formula that I’m going to show you here is uncommon. It is such an important integral that at the end of this chapter we will return to it with some additional analyses.
- 2.
The reason we write a partial derivative inside the integral and a total derivative outside the integral is that the integrand is a function of two variables (t and y) while the integral itself is a function only of t (we’ve ‘integrated out’ the y dependency).
- 3.
The integral in (3.4.8) occurs in a paper by R. M. Dimeo, “Fourier Transform Solution to the Semi-Infinite Resistance Ladder,” American Journal of Physics, July 2000, pp. 669–670, where it is done by contour integration. Our derivation here shows contour integration is actually not necessary. Professor Dimeo states that contour integration is “well within the abilities of the undergraduate physics major,” in agreement with the philosophical position I take in the Preface. Notice that we can use (3.4.8) to derive all sorts of new integrals by differentiation. For example, suppose b = − 1, and so we have \( {\displaystyle {\int}_0^{\uppi}\frac{1}{\mathrm{a}- \cos \left(\mathrm{x}\right)}\mathrm{dx}}=\frac{\uppi}{\sqrt{{\mathrm{a}}^2-1}} \). Then, differentiating both sides with respect to a, we get \( {\displaystyle {\int}_0^{\uppi}\frac{1}{{\left[\mathrm{a}- \cos \left(\mathrm{x}\right)\right]}^2}\mathrm{dx}}=\frac{\uppi \mathrm{a}}{{\left({\mathrm{a}}^2-1\right)}^{3/2}} \). If, for example, a = 5, this new integral is equal to \( \frac{5\uppi}{24^{3/2}}=\frac{5\uppi}{48\sqrt{6}}=0.1335989\dots \). To check, we see that quad(@(x)1./((5-cos(x)).^2),0,pi) = 0.1335989….
- 4.
For n = 1, the recursion gives I1 in terms of I0, where \( {\mathrm{I}}_0={\displaystyle {\int}_0^{\uppi /2}\mathrm{dx}}=\frac{\uppi}{2} \) with no dependency on either a or b. That is, \( \frac{\partial {\mathrm{I}}_0}{\partial \mathrm{a}}=\frac{\partial {\mathrm{I}}_0}{\partial \mathrm{b}}=0 \) and the recursion becomes the useless, indeterminate \( {\mathrm{I}}_1=\frac{0}{0}{\mathrm{I}}_0 \).
- 5.
Uhler was a pioneer in heroic numerical calculation, made all the more impressive in that he worked in the pre-electronic computer days. His major tool was a good set of log tables. I describe his 1921 calculation of, to well over 100 decimal digits, the value of \( {\left(\sqrt{-1}\right)}^{\sqrt{-1}}={\mathrm{e}}^{-\frac{\uppi}{2}} \) in my book An Imaginary Tale: the story of \( \sqrt{-1} \) , Princeton 2010, pp. 235–237.
- 6.
Waldemar Klobus, “Motion On a Vertical Loop with Friction,” American Journal of Physics, September 2011, pp. 913–918.
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Nahin, P.J. (2015). Feynman’s Favorite Trick. In: Inside Interesting Integrals. Undergraduate Lecture Notes in Physics. Springer, New York, NY. https://doi.org/10.1007/978-1-4939-1277-3_3
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DOI: https://doi.org/10.1007/978-1-4939-1277-3_3
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