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2d
revised Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
deleted 157 characters in body
2d
answered Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
Jul
30
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
@tired No, not really.
Jul
30
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
Don't you mean $$ \int_{-\infty}^{\infty} \frac{\tanh^{2n} x}{x^2} \, dx = 2\pi i \sum_{k=0}^{\infty} \underset{z = i \pi/2}{\mathrm{Res}} \frac{\tanh^{2n} z}{(z + i \pi k)^2} ?$$
Jul
29
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
In terms of complex analysis tools, making the substitution $u = \text{arctanh}(x)$ converts it into $$ \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{\pi^{2}+4u^{2}} \, du$$ which can be evaluated by summing the residues in the upper half of the complex plane.
Jul
29
revised Not the toughest integral, not the easiest one
There was a missing exponent.
Jul
29
revised Not the toughest integral, not the easiest one
added 237 characters in body
Jul
29
answered Not the toughest integral, not the easiest one
Jul
29
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
Because that's the only way I see how to show $$ \int_{-\infty}^{\infty} \frac{\tanh^{2n} (x)}{x^{2}} \, dx= \frac{2}{i\pi} \, \underset{z=0}{\mathrm{Res}} \left[ \psi_{1}\left(\tfrac{1}{2} + \tfrac{z}{i\pi} \right) \coth^{2n} (z) \right].$$
Jul
28
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
Specifically I integrated the function $ \psi_{1}\left(\tfrac{1}{2} + \tfrac{1}{i\pi} z\right) \coth^{2n} (z) $, which has poles at $ n \pi i$. The function $\psi_{1}\left(\tfrac{1}{2} + \tfrac{1}{i\pi} z\right)$ has poles along the negative imaginary axis, but they're cancelled by the zeros of $ \coth^{2n} (z)$ if $n >0$.
Jul
28
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
Were you referring to a rectangle in the upper half-plane? I get your result by integrating around a rectangle with vertices at $R - \frac{i \pi}{2}, R + \frac{i \pi}{2}, - R + \frac{i \pi}{2}$, and $-R- \frac{i \pi}{2}$, and then letting $R$ go to infinity.
Jul
28
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
In your second attempt, what contour are you integrating around? My guess is that it is a rectangular contour in the upper half-plane of fixed height.
Jul
27
revised Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$
improved formatting
Jul
23
awarded  Enlightened
Jul
23
awarded  Nice Answer
Jul
23
revised Integration of Bessel function of the second kind
I capitalized the first word in the title.
Jul
23
answered Integration of Bessel function of the second kind
Jul
22
revised A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$
I corrected a misstatement about uniform convergence.
Jul
21
revised Integrating around a pole
I changed the formatting so that it's easier to read.
Jul
20
revised Showing that the Barnes G-function satisfies the functional equation $G(z+1) = \Gamma(z) G(z) $
I gave the question a better title.