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

Question

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?

Answer 1

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.

Answer 2

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 ,

$$z^2=\pi$$

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