# The beat of a different drum

“I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] showed how to differentiate parameters under the integral sign — it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. [If] guys at MIT or Princeton had trouble doing a certain integral, [then] I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” (Richard P. Feynman, “Surely You’re Joking, Mr. Feynman!”, Bantam Book, 1985)

“Feynman’s Trick” is a powerful technique for evaluating nontrivial definite integrals. It is based on Leibniz’s rule (LR-1) which states:

Let $f(x, \beta)$ be a differentiable function in$\beta$ with $\frac{\partial}{\partial \beta}f(x, \beta)$ continuous. Then

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx = \int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

This is how it works in practice:

To evaluate definite integral

$\int\limits_{a}^{b} f(x)\;dx,$

we introduce into integrand $f(x)$ a parameter $\beta$ such that

$f(x) = f(x, \beta)$ when $\beta = \beta_0\quad\quad\quad(1)$

and

$\int\limits_{a}^{b}f(x,\beta)\;dx = f_*$ when $\beta = \beta_*.\quad\quad\quad(2)$

Suppose

$\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x,\beta)\; dx=g(\beta).\quad\quad\quad(3)$

By Leibniz’s rule,

$\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)=\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x,\beta)\; dx\overset{(3)}{=}g(\beta).\quad\quad\quad(4)$

Integrate (4) with respect to $\beta$:

$\int \left(\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx\right) \; d\beta = \int g(\beta)\;d\beta\implies \int\limits_{a}^{b}f(x,\beta)\;dx = G(\beta)+ C\quad(5)$

where $G'(\beta)=g(\beta).$

Let $\beta=\beta_*,$

$\int\limits_{a}^{b}f(x, \beta_*)\;dx \overset{(2)}{=} f_* \overset{(5)}{=} G(\beta_*) + C\implies C=f_*-G(\beta_*).$

Let $\beta = \beta_0,$

$\int\limits_{a}^{b}f(x) \;dx \overset{(1)}{=} \int\limits_{a}^{b}f(x, \beta_0)\;dx\overset{(5)}{=} G(\beta_0) + C.$

And so,

$\int\limits_{a}^{b}f(x) \;dx = G(\beta_0) + f_*-G(\beta_*).$

Now, let’s play “Feynman’s Trick” on definite integral $\int\limits_{0}^{1} \frac{x-1}{\log(x)}\;dx:$

Differentiate $\int\limits_{0}^{1}\frac{x^{\beta}-1}{\log(x)}\;dx$ with respect to $\beta,$ we have

$\frac{d}{d\beta}\int\limits_{0}^{1}\frac{x^\beta-1}{\log(x)}\;dx =\int\limits_{0}^{1}\frac{\partial}{\partial \beta}\frac{x^{\beta-1}}{\log(x)}\;dx=\int\limits_{0}^{1}\frac{x^{\beta}\log(x)}{\log(x)}\;dx=\int\limits_{0}^{1}x^{\beta}\;dx=\frac{x^{\beta+1}}{\beta+1}\bigg|_{0}^{1}=\frac{1}{\beta+1}.$

It means

$\int\limits_{0}^{1}\frac{x^{\beta}-1}{\log(x)}\;dx=\int\frac{1}{\beta+1}\;d\beta = \log(\beta+1)+C\overset{\beta=0}{\implies} 0=\log(0+1) +C \implies C=0.$

Hence,

$\int\limits_{0}^{1}\frac{x^{\beta}-1}{\log(x)}\;dx = \log(\beta+1).$

Let $\beta=1$,

$\int\limits_{0}^{1}\frac{x-1}{\log(x)}\;dx = \log(2).$

Exercise-1 Given $\int\limits_{-\infty}^{\infty}\frac{e^{2x}}{ae^{3x}+b}\;dx = \frac{2\pi}{3\sqrt{3}a^{2/3}b^{1/3}}$ where $a, b >0.$ Show that

$\int\limits_{-\infty}^{\infty}\frac{e^{2x}}{(e^{3x}+1)^2}\;dx = \frac{2\pi}{9\sqrt{3}}.$