Linked Questions

2
votes
1answer
2k views

Infinite Series Problem Using Residues [duplicate]

Show that $$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth\pi a+\frac{1}{2a^2}, a>0$$ I know I must use summation theorem and I calculated the residue which is: $$Res\left(\frac{1}{z^2+...
4
votes
1answer
233 views

The Sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$ [duplicate]

I know how to get the sum of geometric series, but otherwise. How do I get the sum of this series? Thank you. $$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$$
1
vote
2answers
129 views

Prove $\sum_{n\geq1}\frac1{n^2+1}=\frac{\pi\coth\pi-1}2$ [duplicate]

I am trying to prove $$\sum_{n\geq1}\frac1{n^2+1}=\frac{\pi\coth\pi-1}2$$ Letting $$S=\sum_{n\geq1}\frac1{n^2+1}$$ we recall the Fourier series for the exponential function $$e^x=\frac{\sinh\pi}\pi+\...
152
votes
6answers
6k views

Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$

I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$. I found a solution by myself 10 hours after I posted it, here it is: $$f(x)=\...
76
votes
7answers
8k views

How to prove this identity $\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;$?

How to prove this identity? $$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}\;$$ I found the above interesting identity in the book $\bf \pi$ Unleashed. Does anyone knows how to ...
34
votes
5answers
12k views

Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$

This is a homework question whereby I am supposed to evaluate: $$\sum_{n=1}^\infty \frac{1}{n^2 +1}$$ Wolfram Alpha outputs the answer as $$\frac{1}{2}(\pi \coth(\pi) - 1)$$ But I have no idea ...
35
votes
4answers
1k views

Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$.

What is the closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$? We can use Fourier series of $e^{-bx}$ ($|x|<\pi$) to evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{n^2+b^2}$. But this ...
11
votes
6answers
604 views

Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$

How do we evaluate the following series: $$\sum_{n=1}^\infty \frac1{4n^2+2n}$$ I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to ...
11
votes
3answers
403 views

Show that $\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}?$

How may we show that $$\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}\color{red}?\tag1$$ $u={x\over 2}\implies 2du=dx$ $$4\...
2
votes
9answers
347 views

Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?

$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$ I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
10
votes
5answers
571 views

How do I calculate $\sum_{n\geq1}\frac{1}{n^4+1}$?

How do I calculate the following sum $$\sum_{n\geq1}\frac{1}{n^4+1}$$
4
votes
3answers
314 views

Find the closed form for $\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}$

$$I=\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}\tag1$$ $$\ln\left({1+e^{-x}\over 1-e^{-x}}\right)=2\sum_{n=0}^{\infty}{e^{-(2n+1)x}\over ...
5
votes
5answers
1k views

Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$

How to find the Fourier transform of the following function: $$f(x)=\frac{1}{x^2+6x+13}$$
10
votes
4answers
328 views

A curious integral

The following integral has been on my mind for a while $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x \tag{$\dagger$}$$ Let us indicate the integrand as $f(x)=\frac{\sin(x)}{e^x-1}$. The ...
9
votes
2answers
1k views

How many ways to calculate: $\sum_{n=-\infty}^{+\infty}\frac{1}{(u+n)^2}$ where $u \not \in \Bbb{Z}$

Today I have encounter a series: $$\sum_{n=-\infty}^{+\infty}\frac{1}{(u+n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$$ where $u \not \in \Bbb{Z}$ . I have known a method to computer it (by Residue formula): $$...

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