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

and

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

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.

pxchg1200
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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 08 '13 at 12:34
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    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 http://meta.math.stackexchange.com/questions/6424/requests-for-reopen-votes – Gerry Myerson Mar 09 '13 at 01:00
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    What's wrong with this question again? I mean, as opposed to millions of questions being asked and answered here? I'm still learning. – Julien Mar 09 '13 at 01:12
  • well,when you think the problem is too difficult to find a solution,how can you conclude that it's not worth to evaluate it? then closed it? and not effort from proposer? @GerryMyerson, I think try to find a solution instead of close it,then we can make a progress. – pxchg1200 Mar 09 '13 at 01:36
  • @julien, see the recent discussion at http://meta.math.stackexchange.com/questions/8703/academic-integrity (or just see the first comment I left, above). – Gerry Myerson Mar 09 '13 at 01:56
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    @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. – Julien Mar 09 '13 at 02:10
  • @GerryMyerson But even most of these usually get edited and answered. In the present case, there is nothing to say about the format, and the question makes sense. So why picking this one more than another? – Julien Mar 09 '13 at 02:10
  • @julien, first, notice that I didn't --- couldn't --- close the question single-handedly. Four other experienced people agreed with me. Second, notice that the question did get edited and reopened, so, in the end, no harm was done. Third, when you're booked for speeding, "but what about all those other drivers who were speeding?" won't get you off the hook. What's more, at the time I voted to close, this was the only question on the front page that was so poorly presented and hadn't yet attracted any answers. Fourth, a lot of people complain about these questions --- see that meta-thread. – Gerry Myerson Mar 09 '13 at 23:12
  • @GerryMyerson Ok. I was just asking. And I'm not experienced in these matters. – Julien Mar 09 '13 at 23:22
  • 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 08 '13 at 12:59

1 Answers1

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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.

Zander
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