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While going through a list (Problem $17$) of interesting integrals I have come across this one

$$\int_0^1 {\sin \left( {\pi x} \right)} x^x \left( {1 - x} \right)^{1 - x} \mathrm dx = \frac{{\pi e}}{{24}}$$

I have tried integration by parts with $u=x^x \left( {1 - x} \right)^{1 - x}$ and $\mathrm dv=\sin \left( {\pi x} \right)$ but this ended up in and even more complicated integral. To use the series representation of $e^{x\ln(x)}$ and respectively $e^{(1-x)\ln(1-x)}$ does not appeared to be helpful at all even after reshaping it all in the form $(1-x)\left(\frac{x}{1-x}\right)^x$. Then I thought about using Euler's Reflection Formula to get rid of the $\sin(\pi x)$ and work in terms of the Gamma Function instead. Hence this formula only holds for $x\notin\mathbb{Z}$ $-$ and the limits are given by $0$ and $1$ $-$ I guess this is not possible here. My last try was to use the Weierstrass Expansion of the sine function but I guess this is not the right approach either.

Could someone please provide a whole solution since I have no further idea how to deal with this integral? I have searched here on MSE but it does not seem like someone hast asked something like this before. Nevertheless tell me when I have overseen something or when you can link a former question which is helpful for understanding the process of evaluating this definite integral.

Thanks in advance!

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marked as duplicate by Nosrati, mrtaurho, Community Sep 29 '18 at 12:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Nosrati Ah, okay. Thank you. I did not found this. $\endgroup$ – mrtaurho Sep 29 '18 at 12:14
  • $\begingroup$ Sorry about it! $\endgroup$ – Nosrati Sep 29 '18 at 12:15
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    $\begingroup$ An extension is here: math.stackexchange.com/a/1884617/108128 $\endgroup$ – Nosrati Sep 29 '18 at 12:21
  • $\begingroup$ @Nosrati Thank you again. I will look through these two answers to understand the integral completely :) $\endgroup$ – mrtaurho Sep 29 '18 at 12:23