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Evaluate integral $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$ Well,I think we have $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$


$$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$

With such nice result of these integral,why isn't worth to evaluate it?

I found a solution about the second one,but I wonder it will work for the first one 2

Note $$ S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx} $$ Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$ Thus \begin{align} S&=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\ &=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt} \end{align} Due to $$ f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$ Therefore $res(f,0)=-\frac{e}{24}$when $z=0$ with $ \zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$ $$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$ because $$ \{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty $$ Therefore $$ 2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz} $$ gives $$ \int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24} $$

My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.

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Voting to close. No sign of effort from proposer, no attempt to put into context, to explain interest. Needs much improvement to be a suitable question for this site. – Gerry Myerson Mar 8 '13 at 12:34
Result is about 1.15 For extended answer, please provide more of a question (with visible effort to solve it) rather than just pasting a request. – DasKrümelmonster Mar 8 '13 at 12:59
had you put in this new informatiion, together with some indication of why you think these evaluations are correct, the question would not have been closed. You can present a case for reopening at… – Gerry Myerson Mar 9 '13 at 1:00
What's wrong with this question again? I mean, as opposed to millions of questions being asked and answered here? I'm still learning. – 1015 Mar 9 '13 at 1:12
@GerryMyerson Thanks. I'll take the time to read this carefully. Regarding your comment above, I am a little bit confused. There are lots of questions here which do not show any more sign of effort, don't even say please, or thank you. Yet they are answered and no one complains about them. Really a lot. I am just curious why sometimes, apparently randomly, one question is being closed for this reason. Especially in this case where the integral is non trivial. If the question were some trivial linear system awfully formatted, I would understand. – 1015 Mar 9 '13 at 2:10
up vote 2 down vote accepted

Exactly the same method works for the other case. $$ \int_0^1 x^{-x} (1-x)^{x-1}\sin{\pi x} dx = \mathrm{Im}\left[\int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx\right] $$ Write $t=\ln((1-x)/x)$ and $z=t+i\pi$ as you did above to get $$ S = \int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx =\int_{-\infty+i\pi}^{\infty+i\pi} \frac{e^{\frac{z}{1-e^z}}}{1-e^z}dz $$

Then with $$f(z)=\frac{e^{\frac{z}{1-e^z}}}{1-e^z}$$ the only pole is at $z=0$, $res(f,0)=-\frac{1}{e}$ and in the limit $2S = \oint f(z)dz=-2\pi i \cdot res(f,0) = 2\pi i/e$ and your answer follows.

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