Linked Questions

37
votes
5answers
7k views

Riemann zeta function at odd positive integers

Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the ...
40
votes
3answers
8k views

Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$

So I have been trying for a few days to figure out the sum of $$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical analysis and CAS that this sum equals $$ ...
9
votes
5answers
2k views

Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis

How do I compute $$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$ What I am doing is take $$f(z)=\frac{(\log z)^2}{1+z^2}$$ and calculating $\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{...
15
votes
3answers
1k views

Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$

I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+...
6
votes
6answers
354 views

Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$

I recently ran into this series: $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$ Of course this is just a special case of the Beta Dirichlet Function , for $s=3$. I had given the following solution:...
9
votes
3answers
440 views

Calculate $\int_0^1\frac{(\ln x)^2}{\sqrt{x-x^2}}dx=?$

Find the: $$\int_0^1\frac{(\ln x)^2}{\sqrt{x-x^2}}\,dx=\text{?}$$ My Try : $$u=\sqrt{x-x^2} \\ du=\frac{-2x+1}{2\sqrt{x-x^2}}\,dx\\dx=\frac{2\sqrt{x-x^2}}{-2x+1}\,du$$ So we have : $$\int_0^1\...
19
votes
3answers
485 views

Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.

Wolfram$\alpha$ says that we have the following identity $$ \left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}} $$ but, how does one prove such identity?
22
votes
1answer
618 views

Using Fourier Series to compute sums

I have just started learning the basics of Fourier series and have some doubts about it. I am aware that Fourier series can be used to compute infinite sums. For example, $\zeta(2)$ and $\eta(2)$ can ...
3
votes
5answers
208 views

How to solve $\int_0^\infty x^2/(e^{x/2}+e^{-x/2})\text dx$? [duplicate]

I've tried MacLaurin expansion of the integrand, which resulted in $x^2/4 - x^4/16 + x^6/96 + O(x^7)$ When integrated from $0$ to $\infty$ it obviously explodes, but the defined result is ~ $1645$. ...
3
votes
6answers
144 views

Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$

This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up ...
4
votes
1answer
390 views

Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?

Stirling's approximation of the factorial for even numbers is given by $$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$ Further, the Euler numbers grow quite rapidly for large ...
2
votes
2answers
191 views

A closed form for infinite series $ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$

It is well known that $$ \sum _1 ^\infty \frac {1}{n^2} = \frac {\pi ^2}{6}$$ and $$ \sum _1 ^\infty \frac {1}{n^4} = \frac {\pi ^4}{90}$$ We also know that $$ \sum _1 ^\infty \frac {1}{n^3} $$ $$=...
2
votes
1answer
161 views

Elementary method to compute $\zeta (3)$

What is an elementary method to compute $\zeta (3)$ which can be understood by a high school student? I know how to compute $\zeta (2)$ but not $\zeta (3)$. Any help is very much appreciated.
1
vote
2answers
112 views

General P-series rule

This is a p-series: $$\sum_{n=1}^\infty \frac{1}{n^p}$$ There are 2 p-series (to my knowledge) that somehow reach pi: $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ $$\sum_{n=1}^\infty \frac{...
0
votes
1answer
52 views

Find the following improper integral.

Use the function $ f(z) = \dfrac{(\log z)^2}{z^2+1} ; ( |z|>0, - \frac{- \pi}{2} < \arg(z)< \frac{3 \pi}{2})$ to show that $\int_{0}^{\infty} \dfrac{(\ln x)^2}{x^2+1}\, dx = \dfrac{\pi^3}{8}$ ...

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