3
votes
0answers
70 views

An interesting identity involving powers of $\pi$ and alternating zeta series

It happens that, for any $m\geq 1$, $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{1}$$ where $E_{2m}$ is an integer number. My ...
4
votes
0answers
88 views

Ugly-nice double series

I'm trying to evaluate the following ugly double sum (presented in raw notation as used in my calculations): $\sum _{m=1}^{\infty } \sum _{n=1}^{\infty } \frac{4 m \cos \left(\frac{2 \pi m ...
1
vote
2answers
95 views

Infinite sums and integrals using residues

I have no idea how to solve these two, any help? $\mathtt{i)}$ $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{e^{tz}}{\sqrt{z+1}}dz$$ $$ a,t\gt0$$ $\mathtt{ii)}$ $$ \sum_{n=1}^\infty ...
1
vote
1answer
135 views

Evaluating series by contour integration, the residue theorem, and cotangent

I'm trying to understand this section in Tristan Needham's book Visual Complex Analysis about what he says is a standard method for evaluating series via a contour integral. My specific question is ...
23
votes
1answer
541 views

Proving $\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$

Wikipedia informs me that $$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$ I tried considering $f(x,n) = e^{-x n^2}$ so that its ...
5
votes
1answer
111 views

Finding a generalization for $\int_{0}^{\infty}e^{- 3\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(3\pi x)}dx$

$\;\;\;\;$I was reading the introduction of Paul J. Nain's book "Dr. Euler's fabulous formula" where he talks about the sense of beauty in mathematics and quotes the G.N.Watson as saying that a ...
2
votes
1answer
181 views

Need help integrating close contour of $\sec(z)$

I need help integrating this: $$\oint_{|z-1|=1} \sec(z) \, dz $$ From the context of my course, I assume that what I'm suppose to do is expand $\sec(z)$ centered at $z_0=1$ into a Laurent series ...
0
votes
1answer
223 views

The sum of a series to be found by looking at a complex function.

I'm trying to solve the following problem: Find the sum of the series $\sum^{\infty}_{n=1}\frac{(-1)^n}{(2n+1)^3}$ by using the function $f(z)=|(2z+1)^3\sin\pi z|^{-1}.$ I can't see it... I do see ...