I found this integral on Instagram
$$I = \int\limits_0^\infty {{{\left( {\frac{e}{x}} \right)}^x}\Gamma \left( x \right)\sin \left( {2\pi x} \right)dx}$$
My first try is:$$\begin{array}{l}
\displaystyle I = \int\limits_0^\infty {{{\left( {\frac{e}{x}} \right)}^x}\Gamma \left( x \right)\sin \left( {2\pi x} \right)dx} = \sum\limits_{n = 0}^\infty {\int\limits_n^{n + 1} {{{\left( {\frac{e}{x}} \right)}^x}\Gamma \left( x \right)\sin \left( {2\pi x} \right)dx} } \\
\displaystyle {\rm{Let}}:t = x – n \Rightarrow dt = dx \Rightarrow I = \sum\limits_{n = 0}^\infty {\int\limits_0^1 {{{\left( {\frac{e}{{t + n}}} \right)}^{t + n}}\Gamma \left( {t + n} \right)\sin \left( {2\pi t} \right)dt} }
\end{array}$$
But I don't know how to process further from this step.
May I ask for some advices? Thank you very much.
Best Answer
It is possible to calculate the integral exactly without approximations by using the following two results:
$\dfrac{\Gamma(x)}{x^x}=\displaystyle\int_0^{+\infty}t^{x-1}e^{-tx}\mathrm dt\qquad\color{blue}{(1)}$
$\displaystyle\int_{-\infty}^{+\infty}\frac{\mathrm dx}{\big(e^x-1-x\big)^2+4\pi^2}=\dfrac13\qquad\color{blue}{(2)}$
The first result was obtained by Setness Ramesory, whereas the second one was obtained by MathFail ( see here ).
Now we will calculate the integral
$\displaystyle\int_0^{+\infty}\left(\frac ex\right)^x\Gamma(x)\sin(2\pi x)\,\mathrm dx$
by using $\,(1)\,$ and $\,(2)\,$.
$\displaystyle\int_0^{+\infty}\left(\frac ex\right)^x\Gamma(x)\sin(2\pi x)\,\mathrm dx=$
$=\displaystyle\int_0^{+\infty}e^x\left(\int_0^{+\infty}t^{x-1}e^{-tx}\mathrm dt\right)\sin(2\pi x)\,\mathrm dx=$
$=\displaystyle\int_0^{+\infty}\left(\int_0^{+\infty}t^{x-1}e^{(1-t)x}\sin(2\pi x)\,\mathrm dt\right)\mathrm dx=$
$=\displaystyle\int_0^{+\infty}\left(\int_0^{+\infty}\frac1t\left(t\,e^{1-t}\right)^x\sin(2\pi x)\,\mathrm dt\right)\mathrm dx=$
$=\displaystyle\int_0^{+\infty}\left(\int_0^{+\infty}\frac1t\left(t\,e^{1-t}\right)^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt=$
$=\displaystyle\int_0^{+\infty}\frac1t\left(\int_0^{+\infty}\left(t\,e^{1-t}\right)^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt\;.$
Let $\;a=t\,e^{1-t}\,$.
It results that $\;0<a<1\;$ for any $\;t\in(0,+\infty)\setminus\{1\}\,.$
$\displaystyle\int_0^{+\infty}\frac1t\left(\int_0^{+\infty}\left(t\,e^{1-t}\right)^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt=$
$=\displaystyle\int_0^{+\infty}\frac1t\left(\int_0^{+\infty}a^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt=$
$=\displaystyle\int_0^{+\infty}\frac1t\left(\frac{2\pi}{\ln^2\!a+4\pi^2}\right)\mathrm dt=$
$=\displaystyle\int_0^{+\infty}\frac1t\left[\frac{2\pi}{\ln^2\left(t\,e^{1-t}\right)+4\pi^2}\right]\mathrm dt=$
$=\displaystyle\int_0^{+\infty}\frac1t\left[\frac{2\pi}{\big(\ln t+1-t\big)^2+4\pi^2}\right]\mathrm dt=$
$\underset{\overbrace{\text{ by letting }t=e^u\;}}{=}\displaystyle\int_{-\infty}^{+\infty}\frac{2\pi}{\big(u+1-e^u)^2+4\pi^2}\mathrm du=$
$=2\pi\displaystyle\int_{-\infty}^{+\infty}\frac{\mathrm du}{\big(e^u-1-u)^2+4\pi^2}=$
$=\dfrac{2\pi}3\,.$