Evaluate $$\left(\frac{\displaystyle\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}}{\operatorname{coth}(\pi)}\right)^2$$


I saw a very interesting and yet intimidating question on the internet:

Find the value of $$\frac{16\displaystyle\int_0^\pi\int_0^1x^2\cdot\operatorname{sin}(y)\:\:dxdy\:\:\left(\frac{\displaystyle\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}}{\operatorname{coth}(\pi)}\right)^2}{\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}}+5$$

I just know or rather heard that (though I don't know the proof) $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$$ and (I calculated it) $$16\displaystyle\int_0^\pi\int_0^1x^2\cdot\operatorname{sin}(y)\:\:dxdy=\frac{32}{3}$$ but I can't calculate the value of the expression written in the big brackets.

Any help is greatly appreciated.


2 Answers 2


This is going to be tricky, if you don’t know the proof of the Basel problem $\sum_{n=1}^\infty n^{-2}=\frac{\pi^2}{6}$. That said, you can find (very many) proofs of this in many places online, such as this site (see the link)!

The Mittag-Leffler series for the cotangent implies the following identity (by using $i\cot(it)=\coth(t)$, through Euler’s formula): $$\pi\coth(\pi(1))-1/1=\sum_{n=1}^\infty\frac{2(1)}{n^2+(1)^2}=-1+\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}$$That is, $\sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}=\pi\coth(\pi)$. The expression in brackets evaluates to $\pi^2$.

You’re left with: $$5+\frac{16\cdot6}{\pi^2}(\pi^2)\int_0^\pi(\sin y)(1/3)\,\mathrm{d}y=5+64=69$$By integrating first in $x$, then in $y$.

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    $\begingroup$ I’m assuming the final answer was why this overcomplicated problem was popular on the internet… $\endgroup$ Commented Aug 23, 2022 at 13:41
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    $\begingroup$ Person of culture I see @insipidintegrator $\endgroup$
    – Babu
    Commented Aug 23, 2022 at 13:45
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    $\begingroup$ I said so because no normal person who made the problem only for mathematical curiosity, would care to add the “+5” at the end, if it weren’t the objective to have this number as the answer. @TrystwithFreedom $\endgroup$ Commented Aug 23, 2022 at 13:55
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    $\begingroup$ @insipidintegrator A new low in mathematical maturity ; ) $\endgroup$
    – FShrike
    Commented Aug 23, 2022 at 14:11
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    $\begingroup$ @FShrike Ahem. $\endgroup$ Commented Aug 23, 2022 at 14:12

If you enjoy the digamma function

$$S_p=\sum_{n=-p}^{p}\frac{1}{1+n^2}=\sum_{n=-p}^{p}\frac{1}{(n+i)(n-i)}=\frac i 2\sum_{n=-p}^{p}\frac{1}{n+i}-\frac i 2\sum_{n=-p}^{p}\frac{1}{n-i}$$ $$\sum_{n=-p}^{p}\frac{1}{n+i}=-\psi ^{(0)}(p+(1-i))+\psi ^{(0)}(p+(1+i))+i-\psi ^{(0)}(i)+\psi ^{(0)}(-i)$$ $$\sum_{n=-p}^{p}\frac{1}{n-i}=\psi ^{(0)}(p+(1-i))-\psi ^{(0)}(p+(1+i))-i+\psi ^{(0)}(i)-\psi ^{(0)}(-i)$$ $$S_p=-i \psi ^{(0)}(p+(1-i))+i \psi ^{(0)}(p+(1+i))-1-i \psi ^{(0)}(i)+i \psi ^{(0)}(-i)$$ But $$-1-i \psi ^{(0)}(i)+i \psi ^{(0)}(-i)=\pi \coth (\pi )$$

Now, using for large values of $p$ $$\psi ^{(0)}(p+a)=\log(p)+\frac{2 a-1}{2 p}-\frac{6 a^2-6 a+1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ Apply it twice and obtain $$S_p=\pi \coth (\pi )-\frac 2 p+O\left(\frac{1}{p^3}\right)$$


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