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While studying Fourier analysis last semester, I saw an interesting identity:

$$\sum_{n=1}^{\infty}\frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha}$$ whenever $\alpha \in \mathbb{C}\setminus \mathbb{Z}$, which I learned two proofs using Fourier series and residue calculus.

More explicitly, we can deduce the theorem using Fourier series of $f(\theta)=e^{i(\pi - \theta)\alpha}$ on $[0,2\pi]$ or contour integral of the function $g(z)=\frac{\pi}{(z^2-\alpha^2)\tan\pi z}$ along large circles.

But these techniques, as long as I know, wasn't fully developed at Euler's time.

So what was Euler's method to prove this identity? Is there any proof at elementary level?

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    Are you familiar with how he evaluated $\sum_{n \in \Bbb N} \frac{1}{n^2}$? I have no idea, but maybe the same technique works; finding a suitable function and writing it as the infinite product of linear factors, and equating coefficients. – 2017-01-22
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    There were some posts about this series on this site. [Using Approach0](https://approach0.xyz/search/?q=%24%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B1%7D%7Bn%5E2-%5Calpha%5E2%7D%24&p=2) I was able to find [Find the sum of $\sum 1/(k^2 - a^2)$ when $0$\sum_{k=1}^\infty \frac{1}{k^2-L^2}$ and [Calculate $\sum_{k=1}^\infty \frac{1}{k^2-L^2}$ and $\sum_{k=1}^\infty \frac{1}{\left(k-\frac{1}{2}\right)^2-L^2}$](http://math.stackexchange.com/q/1333625). I understand that your question is different - you're asking specifically about ... – 2017-01-22
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    ...Euler's approach. But still links to other posts about the series might be useful for other readers of your question. (BTW I wonder whether the tag ([tag:math-history]) could be appropriate here.) – 2017-01-22
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    Now I have looked also at the question which were listed in the sidebar among related questions. This one might also be of interest: [How did Euler prove the partial fraction expansion of the cotangent function?](http://math.stackexchange.com/q/1849878) – 2017-01-22
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    @MartinSleziak Thanks for your suggestions! – 2017-01-22

2 Answers 2

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According to Wikipedia (https://en.wikipedia.org/wiki/Basel_problem), Euler was the first to give a representation of the sine function as an infinite product: $$(*) \hspace{2cm}\sin (\pi \alpha)=\pi \alpha \prod\limits_{n=1}^{\infty}(\frac{n^2-\alpha^2}{n^2}),$$ which was formally proved by Weierstrass about 100 years later.

Now taking "$\ln$" on by sides of (*) gives $$\ln(\sin (\pi \alpha))=\ln(\pi \alpha)+ \sum \limits_{n=1}^{\infty} \ln (\frac{n^2-\alpha^2}{n^2}),$$ and after taking derivatives on both sides we arrive at

$$\sum_{n=1}^{\infty}\frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha}.$$

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    This sounds really reasonable, even though the completion of this idea was a later development. Thanks! – 2017-01-22
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    @HyJu: Note that Euler's reasoning was probably seriously flawed, but he just was 'lucky'. Even under nice conditions, namely that the zeros are isolated and simple, the zeros only define an analytic function up to a factor of $\exp(g(z))$ for some analytic function $g$. It just happens that $\sin(πz)$ is the product of the linear factors needed to reproduce the zeros, with the $\exp(g(z))$ factor being constant. – 2017-01-22
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    @HyJu: See https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem for more. – 2017-01-22
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    @user21820 Actually I'm studying that part now. Thanks. – 2017-01-22
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For the partial fraction decomposition of the cotangent $$\pi \cot \pi z = \frac{1}{z} + \sum_{n\ge 1}\left( \frac{1}{z-n} + \frac{1}{z+n}\right) = \lim_{k\to\infty} \sum_{n=-k}^k \frac{1}{z-n}$$ hence \begin{align*} \sum_{n=1}^\infty \frac{1}{n^2-\alpha ^2} &= \lim_{k\to\infty}\sum_{n=1}^k \frac{1}{n^2-\alpha^2}\\ &= -\frac{1}{2\alpha}\lim_{k\to\infty} \sum_{n=1}^k \frac{1}{\alpha-n} + \frac{1}{\alpha+n}\\ &= -\frac{1}{2\alpha}\lim_{k\to\infty} \left(-\frac{1}{\alpha} +\sum_{n=-k}^k \frac{1}{\alpha-n}\right)\\ &= -\frac{1}{2\alpha}\left(\pi \cot \pi \alpha - \frac{1}{\alpha}\right)\\ &= \frac{1-\pi \alpha\cot \pi \alpha}{2\alpha^2}\\ &=\frac{1}{2\alpha^2}-\frac{\pi}{2\alpha\tan\pi\alpha} \end{align*}

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    For me the partial fraction decomposition seems already using some aspect of complex analysis. Thanks. – 2017-01-22
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    @HyJu The partial fraction representation can be derived directly from the infinite product representation of the sinc function. – 2017-01-22
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    This development seems completely circular. Starting from the partial fraction representation, one can immediately deduce the coveted identity by simply writing $\cot(\pi \alpha)=\frac{1}{\tan(\pi \alpha)}$ and writing $\frac{1}{z-n}+\frac{1}{z+n}=\frac{2z}{z^2-n^2}$. – 2017-01-22
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    I have a strong suspicion Euler didn't know of the partial fraction decomposition of cotangent, at least not before he has already solved Basel problem. – 2017-01-22