\[ S = \sum_{n=0}^{\infty} \frac{1}{1+n^2} \\ \\ = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) \]
Now , using the unique property of digamma function that is it satisfies, \[ \psi(x+1)-\psi(x)=\frac{1}{x} \]

We get that , sum is equivalent to, \[ 1+ \frac{\psi(1+i) - \psi(1-i)}{2i} \]
Now, again using reflection formula, this can be easily calculated and it equals,

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TopNewestEuler while solving basel problem considered the function \(sin(x)\) as an infinte product as :

\(sin(x)=x\displaystyle \prod _{ n=1 }^{ \infty }{ (1-\dfrac { { x }^{ 2 } }{ { (n\pi ) }^{ 2 } } ) }\)

Taking \(ln()\) both sides we get:

\(ln(sin(x))=ln(x)+\displaystyle \sum _{ n=1 }^{ \infty }{ ln(1-\dfrac { { x }^{ 2 } }{ ({ n\pi })^{ 2 } } ) }\)

Differentiating both sides with respect to \(x\) we get :

\(cot(x)=\dfrac { 1 }{ x } +\displaystyle \sum _{ n=1 }^{ \infty }{ \dfrac { 2x }{ { (n\pi ) }^{ 2 } } \dfrac { 1 }{ (\dfrac { { x }^{ 2 } }{ { (n\pi ) }^{ 2 } } -1) } }\)

\(\Rightarrow cot(x)=\dfrac { 1 }{ x } +\displaystyle \sum _{ n=1 }^{ \infty }{ \dfrac { 2x }{ ({ x }^{ 2 }-{ (n\pi ) }^{ 2 }) } }\)

Putting \(\pi x\) instead of \(x\) we get :

\(cot(\pi x)=\dfrac { 1 }{ \pi x } +\dfrac { 1 }{ \pi } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 2x }{ ({ x }^{ 2 }-{ n }^{ 2 }) } }\)

Multiplying both sides with \(\pi x\) we get :

\(\pi xcot(\pi x)=1+{ 2x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ ({ x }^{ 2 }-{ n }^{ 2 }) } } \)

Also we know that \(icot(ix)=coth(x)\)

Putting \(ix\) in place of \(x\) we get :

\(\pi xcoth(\pi x)=1+2{ x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+{ n }^{ 2 } } } \)

Put \(x=1\) to get :

\(\pi coth(\pi )=1+2\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } \)

\(=2\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } -1\)

\(\Rightarrow \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } =\frac { \pi coth(\pi )+1 }{ 2 } \)

\(\Rightarrow \displaystyle \sum_{n=0}^{\infty}{\frac{1}{n^{2}+1}}=\frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}\)

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That's some identity, isn't it? Nice solution by the way!

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From where you found the question.

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Can you please prove that \(icot(ix) = coth(x)\)

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Easy, we know that :

\(cos(ix) = \frac{{e}^{x}+{e}^{-x}}{2}\)

Also \(sin(ix) = \frac{{e}^{x}-{e}^{x}}{2i}\)

Dividing then we get :

\(cot(ix) = (\frac{{e}^{x}+{e}^{-x}}{{e}^{x}-{e}^{-x}})i\)

Hence finally we have :

\(cot(ix)=icoth(x)\)

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If somebody wants to overkill it, Then

\[ S = \sum_{n=0}^{\infty} \frac{1}{1+n^2} \\ \\ = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) \] Now , using the unique property of digamma function that is it satisfies, \[ \psi(x+1)-\psi(x)=\frac{1}{x} \]

We get that , sum is equivalent to, \[ 1+ \frac{\psi(1+i) - \psi(1-i)}{2i} \] Now, again using reflection formula, this can be easily calculated and it equals,

\[ \frac{1}{2} + \frac{\pi}{2} \coth \pi = \frac{\pi+1}{2} + \frac{\pi}{e^{2\pi}-1}\]

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Are you 16 years old and know digamma function? Wow!

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@Calvin Lin @Jon Haussmann @Michael Mendrin @Steven Zheng @brian charlesworth @Pranav Arora @jatin yadav @Karthik Kannan @Ronak Agarwal Any ideas?

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