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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.

HeroZhang001
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OnTheWay
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1 Answers1

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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 $\;t\in(0,+\infty)\setminus\{1\}\,.$

$\displaystyle\int_0^{+\infty}\frac1t\left(\int_0^{+\infty}\left(t\,e^{1-t}\right)^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt=$

$=\displaystyle\int_0^{+\infty}\frac1t\left(\int_0^{+\infty}a^x\sin(2\pi x)\,\mathrm dx\right)\mathrm dt=$

$=\displaystyle\int_0^{+\infty}\frac1t\left(\frac{2\pi}{\ln^2\!a+4\pi^2}\right)\mathrm dt=$

$=\displaystyle\int_0^{+\infty}\frac1t\left[\frac{2\pi}{\ln^2\left(t\,e^{1-t}\right)+4\pi^2}\right]\mathrm dt=$

$=\displaystyle\int_0^{+\infty}\frac1t\left[\frac{2\pi}{\big(\ln t+1-t\big)^2+4\pi^2}\right]\mathrm dt=$

$\underset{\overbrace{\text{ by letting }t=e^u\;}}{=}\displaystyle\int_{-\infty}^{+\infty}\frac{2\pi}{\big(u+1-e^u)^2+4\pi^2}\mathrm du=$

$=2\pi\displaystyle\int_{-\infty}^{+\infty}\frac{\mathrm du}{\big(e^u-1-u)^2+4\pi^2}=$

$=\dfrac{2\pi}3\,.$

Angelo
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