While I was playing with Wolfram Alpha online calculator I wondered how to calculate and justify an approximation of an integral involving the fractional part function, the exponential and the gamma function $$\int_0^1\left\{\frac{e^x}{\Gamma(x)}\right\}dx. $$ Thus here $\left\{x\right\}$ denotes the fractional part function, see these code

int frac(e^(x)/Gamma(x))dx, from x=0 to 1


plot e^x/Gamma(x), from x=0 to 1

I believe that it is feasible get the good approxiation that I've evoked using (real) analysis, but I am not sure of my strategy.

Question. What is a good method to calculate and justify a good approximation of the integral $$\int_0^1\left\{\frac{e^x}{\Gamma(x)}\right\}dx\,?$$ Many thanks.

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    $\begingroup$ you can use Weierstrass infinite product for $1/\Gamma$ $\endgroup$ – Masacroso Jan 16 '18 at 14:00
  • $\begingroup$ Yours seems a very good hint @Masacroso but I am not able to get the result. My strategy was more simple and without results. $\endgroup$ – user243301 Jan 16 '18 at 14:01
  • $\begingroup$ What exactly was your strategy, just typing it into Wolfram Alpha? $\endgroup$ – Professor Vector Jan 16 '18 at 14:09
  • $\begingroup$ I'm sorry was disconnected @ProfessorVector , my strategy was to combine with the definition of fractional part function, but also with the thought of do a comparison of the numerator and denominator functions, or try to combine with the Fourier expansion of the fractional part. But as I've said I have no results. $\endgroup$ – user243301 Jan 16 '18 at 14:39
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    $\begingroup$ I think a good strategy would be to ignore the fractional part at first and calculate $\displaystyle I_1=\int_0^1\frac{e^x}{\Gamma(x)}\,dx$, first. Then,you subtract the integral over the integer part, being $2-x_1-x_2$, obviously, where $x_k$ is the solution of $\dfrac{e^x}{\Gamma(x)}=k$. All that can be done with standard numerical methods, since $\dfrac{e^x}{\Gamma(x)}$ is a very regular function (entire, actually). The result is $I_1+x_1+x_2-2=0.42384987265144912254672336585425778176$. The only problem with that is: who needs such a result? $\endgroup$ – Professor Vector Jan 16 '18 at 16:30

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