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I was reading this article which derives the value for integral of a negative exponent.

I follow the derivation, from circular symmetry, to integration by substitution, and this is the resulting equation: $$(l(a))^2=\pi\int_0^\infty e^{(-ax)} \, dx=\frac \pi a\tag1$$ (where $l(a)$ is some function on $a$.)

And we take the $\sqrt{}$ of equation (1), which produces: $$l(a)=\int_{-\infty}^\infty e^{(-ax^2)} \, dx=\sqrt{\frac \pi a} \text{ for } \operatorname{Re}(a)>0\tag2$$

I understand that the bound became negative infinity because taking the square root gives negative solutions.

However, I don't understand why the exponential factor became $x^2$, and I have no idea where the $\pi$ from equation (1) went. Why did the integral eat the $\pi$?

Au101
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Zhengqun Koo
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  • That's not how it works. It becomes $\sqrt{\int_0^\infty e^{-ax}dx}$ – Simply Beautiful Art Nov 20 '16 at 13:49
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    Are you familiar with polar coordinates? That's where the $\pi$ went. – Simply Beautiful Art Nov 20 '16 at 13:54
  • Notice they didn't take the square root of $(1)$. Instead, they squared $(2)$ and turned it into $(1)$. – Simply Beautiful Art Nov 20 '16 at 13:56
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    "Taking a square root" does not "give negative solutions". Even the thing that you mean by that phrase has nothing to do with the lower bound of the integral becoming $-\infty$. – Eric Towers Nov 20 '16 at 13:59
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    It's really a shame that the equations in the linked article were put in such low-resolution images. It becomes almost impossible to tell what some of the individual letters in each equation are supposed to be. For example, what you read as lower-case L (ell) was meant to be an upper-case I (eye). – David K Nov 20 '16 at 16:34
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    When I saw the title, I honestly thought there was going to be a joke in the answer. – Skeleton Bow Nov 20 '16 at 21:16
  • Thanks so much guys! My integral variable for equation (1) is wrong, it should be $u$ (where $u=r^2=x^2+y^2$), instead of $x$ (thinking it was $x$ had me conflating (1) with the earlier integral of $\int_0^\infty{e^{-ax}}dx$, when both integrals are entirely different). I totally blame this on the font haha. Also, the bounds changed from $-\infty\ to\ 0$ because we switched to integration in terms of $radius\ >\ 0$ (not because of square root). Thank you all once again for the clarification. – Zhengqun Koo Nov 27 '16 at 16:26
  • Possible duplicate of Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$ – Nosrati Sep 15 '18 at 04:04

1 Answers1

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I think you're conflating two different things in that article.

First, they discuss the integral $\int_0^\infty e^{-a x}\,dx$ and show that it equals $1/a$. This follows by recognizing $\frac{1}{a}e^{-a x}$ as an antiderivative.

Second, they then want to separately compute the integral $I(a)=\int_{-\infty}^\infty e^{-a x^2}\,dx$. To proceed on this second integral, they write $$I(a)^2 = \int_{-\infty}^\infty e^{-a x^2}\,dx\int_{-\infty}^\infty e^{-a y^2}\,dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-a (x^2+y^2)}\,dxdy.$$ To proceed further, they note that the rotational symmetry permits the integration over all $x,y$ to be converted into an integral over $r=\sqrt{x^2+y^2}$ with weight $2\pi r\,dr$. Hence $$~~I(a)^2 = \int_0^\infty 2\pi r e^{-a r^2}\,dr=2\pi\underbrace{\int_0^\infty\frac12e^{-au}du}_{\large u=r^2,\ du=2rdr}=2\pi\left[\frac1{2a} e^{-ar^2}\right]_0^\infty = \frac{\pi}{a}~~$$ which upon taking the square root gives the desired result.

Viktor Toth
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Semiclassical
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