We know that $\sum_{k=1}^\infty \frac{1}{k^2+1} = \frac{1}{2}(\pi \coth(\pi) - 1)$. Now, how do we calculate the series $\sum_{k=1}^\infty \frac{1}{(k + x)^2+1}$ for $x\geq 0$?
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Possible duplicate of Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$ – Jan 03 '18 at 15:33
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I won't say it's a duplicate, rather a generalization. – Olivier Oloa Jan 03 '18 at 15:43
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Thank you so much. Olivier Oloa – koohyar eslami Oct 07 '18 at 08:51
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One may write $$ \begin{align} \frac{2i}{(k + x)^2+1}&= \frac{1}{k+x-i}-\frac{1}{k+x+i} =\!\left(\frac{1}{k}-\frac{1}{k+x+i}\right)-\left(\frac{1}{k}-\frac{1}{k+x-i}\right) \end{align} $$ yielding
$$ \sum_{k=1}^\infty \frac{1}{(k + x)^2+1}=\frac1{2i}\psi(x+1+i)-\frac1{2i}\psi(x+1-i),\qquad \text{Re}(x+1)>-1,\tag 1 $$
where we have used the digamma function which satisfy $$ \psi(z+1)=-\gamma+\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+z} \right),\quad \text{Re}z>-1.\tag2 $$ For example, by putting $x=\frac12$ in $(1)$, using special values of $\psi$, one gets
$$ \sum_{k=1}^\infty \frac{1}{\left(k + \frac12\right)^2+1}=\color{blue}{-\frac45+\frac \pi2 \tanh (\pi)}.\tag3 $$
Olivier Oloa
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https://math.stackexchange.com/questions/736860/find-the-infinite-sum-of-the-series-sum-n-1-infty-frac1n2-1 – Jan 03 '18 at 15:33
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@ Olivier Oloa, Hi, I have question that is : I work on a subject which required your closed form mentioned in (1). I use (1) in some inequalities and I want to know that is (1) real or not because of (i)? thanks. – soodehMehboodi Feb 17 '19 at 18:21
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@soodehMehboodi If $x$ is a real number, then both sides of $(1)$ are real. Observe that, in this case $$\frac1{2i}\psi(x+1+i)-\frac1{2i}\psi(x+1-i)=\Im\left[\psi(x+1+i)\right]$$ which is a real number, as $x$ is a real number. – Olivier Oloa Feb 18 '19 at 10:26
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@ Olivier Oloa, Excuse me, would you tell me : how can you prove that (1) is real. moreover, what is the function on the right hand of identity mentioned in your comment? thanks – soodehMehboodi Feb 18 '19 at 11:13
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@soodehMehboodi If $x$ is a real number, then each $\displaystyle \frac{1}{(k + x)^2+1}$, $k=1,2,\cdots,$ is a real number, then $\displaystyle \sum_{k=1}^\infty \frac{1}{(k + x)^2+1}$ is a real number, as a sum of real numbers. The related function is the digamma function $\psi$: https://en.wikipedia.org/wiki/Digamma_function. – Olivier Oloa Feb 18 '19 at 16:18
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