# mrtaurho

Germany

"The mathematician does not study pure mathematics because it is useful; he studies it because he delights in it and he delights in it because it is beautiful" - Georg Cantor

Member of the Integral Fan Club ^^:@Zacky@Kemono Chen @clathratus @omegadot@ComplexYetTrivial @Sangchul Lee@Olivier Oloa

Contact: mrtaurho[at]gmail[dot]com

My favorite Theorem so far:

Ramanujan's Master Theorem

Let $$f(x)$$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\phi(-s)$$

Some contributions I am proud of:

• How to show that $\prod\limits_{r=1}^{n}\Gamma{ \left({\frac {r}{n+1}}\right)}={\sqrt {\frac {(2\pi )^{n}}{n+1}}}$?
• A Series For the Golden Ratio
• Proving $\Im\operatorname{Li}_2(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1)$
• Is $\int\limits_0^\infty\frac{\sin y}{y^{s+1}}dy=-\Gamma(-s)\sin(\frac{\pi s}{2})$ for $\operatorname{Re}(s)\in (-1,0)$ obvious?
• $\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$
• Integral of $\ln(\tanh(x))$
• Integral $\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$
• Why is Catalan's constant $G$ important?

And even some of my own question I would call interesting $$($$^^$$)$$:

• Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$
• Show that $\sum\limits_{n=1}^{\infty}\frac1{n^2}=\sum\limits_{n=1}^{\infty}\frac3{n^2\binom{2n}n}$ without actually evaluating both series
• Evaluate$\int\limits_0^1 [\log(x)\log(1-x)+\operatorname{Li}_2(x)]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$
• Show that $\int_0^1 \frac{\ln(1+x)}x\mathrm dx=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx$ without actually evaluating both integrals
Top Questions

## Show that $\sum\limits_{n=1}^{\infty}\frac1{n^2}=\sum\limits_{n=1}^{\infty}\frac3{n^2\binom{2n}n}$ without actually evaluating both series

asked Dec 24 '18 at 12:46

## On the integral $I(a)=\int_0^1\frac{\log(a+t^2)}{1+t^2}\mathrm dt$

asked Nov 11 '18 at 18:57