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I've been learning a technique that Feynman describes in some of his books to integrate. The source can be found here:

http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf

The first few examples are integrals in $x$ and $y$ variables, and I can't see a good way to simplify them using differentiation, particularly the example:

$$\int^\infty_0 \exp\left(\frac{-x^2}{y^2}-y^2\right) dx$$

Zev Chonoles
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Ryan Joseph
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    Take out $exp(-y^2)$ and integrate directly, it is gaussian. – Rogelio Molina May 22 '15 at 19:01
  • I see another example in there that is just $\int_0^\infty \exp(-x^2)dx$. Would love to see how this technique can be used to provide an alternate derivation. – muaddib May 22 '15 at 19:12
  • is there any way that it is "dy" instead of"dx"? – tired May 22 '15 at 19:26
  • If you scroll down to the bottom of the pdf linked it's the first example, unless they made a mistake, it's right. – Ryan Joseph May 22 '15 at 19:50
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    Just for note: There is similar, but harder problem, when you have $dy$ instead of $dx$. If you want to solve that problem (where $x$ is parameter, say $x>0$ (but you can do all for $x \in \mathbb{R}$)), first find $I'(x)$ and then use substitution $y \mapsto \frac{x}{t}$ for $I(x)$, where $I(x)$ is your integral. You should end with something like $I'(x)+2I(x)=0$, that is $I(x)=Ce^{-2x}$ (you can find that $C=\frac{\sqrt{\pi}}{2}$ from $I(0)=\lim_{x \to 0^+} I(x)$). – Cortizol May 22 '15 at 20:19
  • Related to Cortizol's comment and Dr. MV's answer see this thread. – Raymond Manzoni May 22 '15 at 21:54
  • This page might help https://highvoltagemath.wixsite.com/highvoltagemath/the-gaussian-integral-take-2 – Highvoltagemath Aug 22 '19 at 13:02

2 Answers2

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Suppose the integral were $I=\int_0^{\infty} e^{-y^2-\frac{x^2}{y^2}}dy$. Then we note that $y^2+\frac{x^2}{y^2}=\left(y-\frac{|x|}{y}\right)^2+2|x|$.

Thus, we have

$$I=e^{-2|x|}\int_0^{\infty}e^{-\left(y-\frac{|x|}{y}\right)^2}dy \tag 1$$

Now, substitute $y\to |x|/y$ so that $dy\to -|x|dy/y^2$. Then,

$$I=e^{-2|x|}\int_0^{\infty}\frac{|x|}{y^2}e^{-\left(y-\frac{|x|}{y}\right)^2}dy \tag 2$$

If we add $(1)$ and (2), we find

$$\begin{align} I&=\frac12\,e^{-2|x|}\int_0^{\infty}\left(1+\frac{|x|}{y^2}\right)e^{-\left(y-\frac{|x|}{y}\right)^2}dy \\\\ &=\frac12\,e^{-2|x|}\int_{-\infty}^{\infty}e^{-y^2}dy\\\\ &=e^{-2|x|}\frac{\sqrt{\pi}}{2} \end{align}$$

So, while not quite a "Feynmann" trick, it is an effective way of evaluation.

Mark Viola
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I looked through the paper and see that the "Feynman Technique" is really just a clever application of Leibniz's rule for taking derivatives under integrals. As is pointed out in the comments, the interesting part of this is evaluating $$\int_0^\infty e^{-x^2} dx$$ using the technique. The paper is a bit misleading for this one since all the other examples added a function of $b$ into the integrand, but that doesn't seem to be the right way to do this one. Instead consider: $$I(b) = \left(\int_0^b e^{-x^2} dx\right)^2$$ Ultimately you want to evaluate that at $b = \infty$ and take its square root.

The full derivation of the result can be found here. That paper discusses this in terms of the Leibniz rule and gives several other interesting derivations of different integrals.

muaddib
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