The result $$ \int_{\infty}^\infty\frac{dx}{\left(e^xx+1\right)^2+\pi^2}=\frac{1}{2}$$ holds and can be easily extracted via residues.
Use the above to prove that: $$ \int_0^\infty x^{x} e^{x}\, \Gamma(x) \sin (\pi x)\, dx= \frac {\pi}{ 2}$$ whereas \( \Gamma \) is the Gamma function defined as \(\displaystyle \Gamma(x)=\int_0^\infty t^{x1}e^{t} \, dx \) and \(x^{x} \) is the iterated function.
Welcome to mathimatikoi.org forum; Enjoy your visit here.
Gamma, trigonometric and iterated integral
 Tolaso J Kos
 Administration team
 Articles: 2
 Posts: 855
 Joined: Sat Nov 07, 2015 6:12 pm
 Location: Larisa
 Contact:
Gamma, trigonometric and iterated integral
Imagination is much more important than knowledge.

 Articles: 0
 Posts: 10
 Joined: Sun Sep 04, 2016 5:08 am
Re: Gamma, trigonometric and iterated integral
This is very beautiful, can you give me some hint ?
Civil Engineer