Proving ∑∞n=−∞e−πn2=4√πΓ(34)\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}

Wikipedia informs me that

S=ϑ(0;i)=n=eπn2=4πΓ(34)

I tried considering f(x,n)=exn2 so that its Mellin transform becomes Mx(f)=n2zΓ(z) so inverting and summing

12(S1)=n=1f(π,n)=n=112πic+icin2zΓ(z)πzdz=12πic+iciζ(2z)Γ(z)πzdz

However, this last integral (whose integrand has poles at z=0,12 with respective residues of 12 and 12) is hard to evaluate due to the behavior of the function as (z)± which makes a classic infinite contour over the entire left/right plane impossible.

How does one go about evaluating this sum?

Answer

This one is a direct evaluation of elliptic integrals. Jacobi’s theta function ϑ3(q) is defined via the equation ϑ3(q)=n=qn2 Let 0<k<1 and k=1k2 and we define elliptic integrals K,K via K(k)=π/20dx1k2sin2x,K=K(k),K=K(k) Then it is almost a miracle that we can get k in terms of K,K via the variable q=eπK/K using equations k=ϑ22(q)ϑ23(q) where ϑ2(q) is another theta function of Jacobi defined by ϑ2(q)=n=q(n+(1/2))2 Also the function ϑ3(q) is directly related to K via ϑ3(q)=2Kπ The proofs of (3) and (5) are given in the linked post on my blog.


The sum in the question is ϑ3(eπ) so that we have q=eπ. This implies that K/K=1 so that k=k and from k2+k2=1 we get k2=1/2. And then ϑ3(q)=2Kπ=2πΓ2(1/4)4π=Γ(1/4)π3/42 Now using Γ(1/4)Γ(3/4)=π/sin(π/4)=π2 we get n=eπn2=ϑ3(eπ)=4πΓ(3/4) The value of K=K(1/2) in terms of Γ(1/4) is evaluated in this answer.

Attribution
Source : Link , Question Author : Argon , Answer Author : Paramanand Singh

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