# Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$

Prove that $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$.

Here is my attempted solution:

Define $a:=\sqrt{\pi}e^{\frac{\pi i}{4}}$ and let $f(z) = \frac{e^{-z^2}}{1+e^{-2az}}$.

Note that $a^2=\pi i$.

Now $f(z)$ has poles of order 1 at $(k+\frac{1}{2})a$ for all $k\in\mathbb{Z}$. Thus using the Residue Theorem and l'Hôpital's Rule:

$$\lim_{z\rightarrow (k+\frac{1}{2})a}\frac{(z-(k+\frac{1}{2})a)e^{-z^2}}{1+e^{-2az}}=\lim_{z\rightarrow (k+\frac{1}{2})a}\frac{\frac{d}{dz}(z-(k+\frac{1}{2})a)e^{-z^2}}{\frac{d}{dz}1+e^{-2az}}=\frac{e^{-(k+\frac{1}{2})^2a^2}}{-2ae^{-2(k+\frac{1}{2})a^2}}=\frac{e^{-k^2a^2-ka^2-\frac{a^2}{4}}}{-2ae^{-2ka^2-a^2}}=\frac{e^{-\pi ik^2-\pi ik-\frac{\pi i}{4}}}{-2ae^{-2\pi ik-\pi i}}=\frac{e^{\frac{-\pi i}{4}}}{2\sqrt{\pi}e^{\frac{\pi i}{4}}}=\frac{1}{2\sqrt{\pi}i}$$

It can be shown that $f(z)-f(z+a)=e^{-z^2}$ (I wont prove this but it's most definitely true). I'm going to integrate this function around the contour which is a rhombus slanting to the right whose bottom two corners lie at $-R$ and $R$. Thus we have:

$$\lim_{R\rightarrow\infty}\bigg[\int_{-R}^{R}f+\int_{R+at}f+\int_{R+a}^{-R+a}f+\int_{R+a(1-t)}f\bigg]=\sum_{k\geq 0}\sqrt{\pi}$$ for $0\leq t\leq 1$.

Making the substitutions $u=z-a$ for the third integral and then $v=1-t$ for the fourth, and then taking $R\rightarrow\infty$, we obtain:

$$\int_{-\infty}^{\infty}e^{-z^2}dz=\sum_{k\geq 0}\sqrt{\pi}$$

Whew! Ok so clearly all I really want is just one $\sqrt{\pi}$, not an infinite number of them. But when I take $R\rightarrow\infty$ my rhombus contains all the poles in the upper-half plane. Can anyone tell where I went wrong?

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$$1+e^{-2az}=0\Longleftrightarrow e^{-2az}=-1\Longleftrightarrow 2az=\frac{2k+1}{2}\pi i\Longleftrightarrow z=\frac{2k+1}{4a}\pi i=:w_k\;,\;\;k\in\Bbb Z$$ so I don't understand where did you get your function's poles from...or, of course, I missed something. –  DonAntonio Feb 17 '13 at 2:38
The only thing different between mine and yours is you have an extra factor of 2 in the denominator, and I think that might have been a mistake on your part. –  Tyroshi pleasure barge Feb 17 '13 at 2:43
I don't think that's the only difference: you also have $\,a\,$ multiplying, I have it dividing... –  DonAntonio Feb 17 '13 at 2:44
Yes but $a=\sqrt{\pi}e^{\frac{\pi i}{4}}$, so bringing it up to the numerator and making the powers negative simplifies to $a$ up above, since $i=e^{\frac{\pi i}{2}}$. Or I guess simpler would be just to refer to my note that $\pi i=a^2$. –  Tyroshi pleasure barge Feb 17 '13 at 2:45
See this also: math.stackexchange.com/questions/34767/… –  Byron Schmuland Feb 17 '13 at 3:19

## 2 Answers

This is a proof which is not at all 'Complex Analytic' but is very elementary so I thought of sharing it as an answer to this question.

Let $\int_{-\infty}^{\infty}e^{-x^2}dx=z$. Clearly $z=2\int_{0}^{\infty}e^{-x^2}dx$. Now as exponential function is positive so we have, $z\geq 0$(It is rather strict)

Then we have $\int_{-\infty}^{\infty}e^{-x^2}dx.\int_{-\infty}^{\infty}e^{-y^2}dy=z^2$

We have,

$$z^2=\int_{-\infty}^{\infty}e^{-x^2}dx.\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2}e^{-y^2}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x+y)^2}dxdy$$

Changing into the polar co-ordinates we have,

$$x=r\cos \theta,y=r \sin \theta$$ $$\Rightarrow dxdy=rdrd\theta$$

Replacing in the above integral we have,

$$\int_{-\pi}^{\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta$$

$$= \frac{1}{2}\int_{-\pi}^{\pi}\int_{0}^{\infty}e^{-y}dyd\theta$$ $$= \frac{1}{2}\int_{-\pi}^{\pi}d\theta$$ $$=\pi$$

So we have ,

$$z^2=\pi$$

$$\Rightarrow z=\sqrt{\pi}$$

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What do you mean by "exponential function positive"? Complex function can be neither positive nor negative, since positivity is something related to real numbers only. Also, what exactly do you mean by introducing polar coordinates for complex number $z = x + iy$? Anyway, point is proof you provided is for real numbers. –  Kaster Feb 17 '13 at 5:15
Yes the proof is for real numbers. –  Abhra Abir Kundu Feb 17 '13 at 5:24
Right. But it supposed to be proved for complex numbers :) –  Kaster Feb 17 '13 at 5:34

Ok everyone you can stop furiously scribbling to check all my calculations. Byron Schmuland sent me to a post which has this same solution and it made me realize my mistake:

Taking $R\rightarrow\infty$ only extends this rhombus in the horizontal directions! So I don't pick up anymore singularities in the limit. So there it is.

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