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For $n\geqslant m>1$, the integral $$I_{n,m}:=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx$$ converges. If $m$ and $n$ are both even or both odd, we can use the residue theorem to easily evaluate it …

Inspired by the well known $$\int_0^1\frac{\ln(1-x)\ln x}x\mathrm dx=\zeta(3)$$ and the integral given here (writing $\zeta_r:=\zeta(r)$ for easier reading)$$\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx= …

For $n\ge m\ge 2$, define $$I(n,m):= \int _0^\infty\dfrac{\text{arcsinh}^nx}{x^m}dx$$ Computer algebra systems say that the indefinite integral can be expressed in terms of polylog functions (of rapid …

A reflection formula for the Hurwitz zeta function, which does not seem to be well known, uses half of the polynomials generated by $\frac{1}{-1+\sqrt{t-1}\cot(\sqrt{t-1}u)}$. (Look at the sections "f …

It is well-known that the Euler constant $$\gamma=\lim\limits_{n\to \infty}\biggl( \sum\limits_{k\le n}\dfrac{1}{k}-\ln{n}\biggr ) $$ has a bunch of integral representations, e.g. $$\gamma=-\int\limit …

The beta function $B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$ can be written for $\Re x, \Re y > 0$ symmetrically as $$ B(x,y) = \int_{-\frac12}^{\frac12}(\frac12-t)^{x-1}(\frac12+t)^{y-1}\,dt. …

It is well-known since Euler that the Generalized harmonic numbers, defined for $n\in\mathbb N$ by $$H_n^{(r)}=\sum_{k=1}^n\frac1{k^r},$$ can be naturally extended for non integer $n$ in terms of poly …

For naturals $n\ge m$, define $$I(n,m):=\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$$ with $\text{arcsinh}\ x=\ln(x+\sqrt{1+x^2} )$, so $\text{arcsinh} \frac12=\ln \frac{\sqrt{5}+1}2 $. Is it poss …

This is inspired by the same beautiful integral expression for $\zeta(3)$ as this question, but goes in a slightly different direction. Writing the original integral in the form $$\int_0^1\frac{x(1-x) …