TITLE: Integrate $\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$
QUESTION [12 upvotes]: 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

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.

REPLY [6 votes]: 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.