Inspired by this recently closed question, I’m curious whether there’s a way to do the Gaussian integral using techniques in complex analysis such as contour integrals.

I am aware of the calculation using polar coordinates and have seen other derivations. But I don’t think I’ve ever seen it done with methods from complex analysis. I am ignorant enough about complex analysis to believe it can somehow be done without knowing how it would be done.

**Answer**

What follows is a list of solutions that I enjoy, and use complex analysis either implicitly or explicitly. I will update the list as I come up with more. (**Note:** **Solution 4** is my favorite, and is completely complex analysis oriented. I also quite like **Solution 6.**)

First, let u=x2, du=2xdx. Then our integral becomes ∫∞−∞e−x2dx=∫∞0u−12e−udu=Γ(12). where Γ(s) is the Gamma function.

**Solution 1:**

Since Γ(1−s)Γ(s)=πsinπs for all complex s, we conclude Γ(12)=√π.

**Solution 2:** Recall the Beta function, B(x,y)=∫10tx−1(1−t)y−1dt=Γ(x)Γ(y)Γ(x+y). Setting x=y=12 we have

(Γ(12))2=∫101√t(1−t)dt.

To evaluate this, set t=sin2(x) to find (Γ(12))2=∫π202sinxcosxsinxcosxdt=π.

Alternatively, we could evaluate the last integral by choosing branch’s such that the integrand is analytic on C−[0,1] and then integrating around this cut. (The residue then comes from the residue at infinity)

**Solution 3:** Setting s=12 in the duplication formula, Γ(s)Γ(s+12)=√π21−2sΓ(2s), yields Γ(12)=√π.

**Solution 4: My personal favorite:**

Recall the functional equation for the zeta function, namely that π−z2Γ(z2)ζ(z)=π−1−z2Γ(1−z2)ζ(1−z). Taking the limit as z→1, we know that ζ(z)∼1z−1 and Γ(1−z2)∼21(z−1) so that we must have the equality π−12Γ(12)=2ζ(0). By taking the limit in the right half plane as s→0 using the identity ζ(s)=ss−1−s∫∞1{u}u−sdu, which holds for σ>0, we can find that ζ(0)=12. (notice the pole/zero cancellation). Consequently Γ(12)=√π.

**Solution 5:** From complex integration, for a,b>0 we have the identity ∫∞−∞(1−ix)−a(1+ix)−bdx=22−a−bπΓ(a+b−1)Γ(a)Γ(b). Set a=12,b=32 to find that

∫∞−∞1−ix(1+x2)32dx=2πΓ(12)2. Hence ∫∞01(1+x2)32dx=πΓ(12)2. Since the integrand on left hand side has anti derivative x√x2+1+C, it follows that the integral is 1 and henceΓ(12)=√π.

**Solution 6:** More with the Beta function. Consider the Mellin Transform M(1(1+t)a)(z):=∫∞0tz−1(1+t)adt=B(a−z,z). The last equality follows by substituting v=11+t, and then rewriting the integral as ∫10va−z−1(1−v)z−1dv. Now, plug in a=1 and z=12 to get ∫∞01√t(1+t)dt=Γ(12)2 and then let t=x2 to find

2∫∞011+x2dx=π=Γ(12)2.

**Solution 7:** We can also prove the result by using Stirling’s formula. Admittedly, this isn’t really using complex analysis, but I find it interesting.

Since zΓ(z)=Γ(z+1) we see that

Γ(n+12)=Γ(12)⋅(12)(32)⋯(2n−12)=Γ(12)((2n)!n!4n)=Γ(12)(2nn)n!4n.

By Stirling’s formula,

\binom{2n}{n}\frac{1}{4^{n}}\sim\frac{1}{\sqrt{\pi n}}\ \text{as}\ n\rightarrow\infty

and

\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\sim\frac{1}{\sqrt{ne}}\frac{\left(n+\frac{1}{2}\right)^{n}}{n^{n}}.

Using the fact that \lim_{n\rightarrow\infty}\left(1+\frac{a}{n}\right)^{n}=e^{a}, it then follows that

\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\sim\frac{1}{\sqrt{n}}.

Consequently, taking the limit as n\rightarrow\infty in the formula \Gamma\left(\frac{1}{2}\right)=\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\frac{4^{n}}{\binom{2n}{n}} yields \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.

Hope that helps,

**Remark:** All of the formulas used here can be proven without use the fact that \Gamma(1/2)=\sqrt{\pi}, so that none of these are cyclic. This is mainly worth pointing out for 4.

**Edit:** I put what were solutions 2 and 3 together since they were not different.

**Attribution***Source : Link , Question Author : JasonMond , Answer Author : Did*