# Evaluate $\sum\limits_{n=0}^\infty \frac{1}{4n^2+1}$ by using complex contour integration

How do you evaluate $\displaystyle\sum\limits_{n=0}^\infty \frac{1}{4n^2+1}$ by using complex contour integration?

I'm trying to attempt this question by considering the integral of some function about a square in the complex plane, whose residues at each singularity on the real axis evaluate to $\large\frac{1}{4k^2+1}$ for all integers $k$. Maybe a function similar to $$\frac {\cot \pi z}{4z^2+1}$$

Maybe then define a square centred on the origin with sides of length $2N+1$ then letting $N \to \infty$ we can split up the integral to evaluate the sum of the residues? Which would be our summation. Sorry if this is poorly explained, but as I say, i'm having trouble understanding this, so i'm not that sure myself, I know it's possible though!

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See math.stackexchange.com/questions/8337/… for a similar computation. – David Speyer Nov 17 '11 at 19:18

If you choose to evaluate this using a contour it will not just be similar, but identical to the contour for $\cot(\pi z)$ with a variable change.
By contour integration we have that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty \frac{1}{z^2-n^2}.$$ Let $z=\frac{i}{2}$, then this is $$\frac{\pi i}{2}\cot\left(\frac{\pi i}{2}\right)=1+2\sum_{n=1}^\infty \frac{1}{4n^2+1}.$$ Since $\coth(z)=i\cot(iz)$ we conclude that $$\sum_{n=0}^\infty \frac{1}{4n^2+1}=\frac{1}{2}+\frac{\pi}{4}\coth\left(\frac{\pi}{2}\right).$$ as desired.
If you want to do the variable change before the contour integration, just look at $z\rightarrow \frac{iz}{2}$. Then we are integrating the function $\pi z \coth \left( \frac{\pi z}{2}\right)$, which really the same function as $\pi z\cot\left(\pi z\right)$.
Oh, don't you mean $\coth(z)=i\cot(iz)$, and do you not need to add $1$ to your final statement to take into account the fact the summation starts from zero. – Freeman Nov 17 '11 at 20:13