### Gamma, trigonometric and iterated integral

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**Thu Jul 14, 2016 7:04 am**The result $$ \int_{-\infty}^\infty\frac{dx}{\left(e^x-x+1\right)^2+\pi^2}=\frac{1}{2}$$ 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 \(\displaystyle \Gamma(x)=\int_0^\infty t^{x-1}e^{-t} \, dx \) and \(x^{-x} \) is the iterated function.

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 \(\displaystyle \Gamma(x)=\int_0^\infty t^{x-1}e^{-t} \, dx \) and \(x^{-x} \) is the iterated function.