Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $ I've found here the following integral.
$$I = \int_{0}^{1}\sin{(\pi (1-x))}x^x(1-x)^{1-x}\,dx=\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}$$
I've never seen it before and I also didn't find the evaluation on math.se. How could we verify it?
If it is a well-known integral, then could you give a reference?
 A: This one can be done with "residue at infinity" calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration . 
First, we use $z^z = \exp ( z \log z )$ where $\log z$ is defined for $-\pi\leq \arg z < \pi$. 
For $(1-z)^{1-z} = \exp ( (1-z)\log (1-z) )$, we use $\log (1-z)$ defined for $0\leq \arg(1-z) <2\pi$. 
Then, let  $f(z)= \exp( i\pi z + z \log z + (1-z) \log (1-z) )$. 
As shown in the Ex VI in the wikipedia link, we can prove that $f$ is continuous on $(-\infty, 0)$ and $(1,\infty)$, so that the cut of $f(z)$ is $[0,1]$. 
We use the contour: (consisted of upper segment: slightly above $[0,1]$, lower segment: slightly below $[0,1]$, circle of small radius enclosing $0$, and circle of small radius enclosing $1$, that looks like a dumbbell having knobs at $0$ and $1$, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)
On the upper segment, the function $f$ gives, for $0\leq r \leq 1$, 
$$\exp(i\pi r) r^r (1-r)^{1-r} \exp( (1-r) 2\pi i ).$$ 
On the lower segment, the function $f$ gives, for $0\leq r \leq 1$, 
$$\exp(i\pi r) r^r(1-r)^{1-r}. $$
Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero. 
Thus, the integral of $f(z)$ over the contour, is the integral over the upper and lower segments, which contribute to
$$\int_0^1 \exp(i\pi r) r^r (1-r)^{1-r} dr - \int_0^1 \exp(-i\pi r) r^r(1-r)^{1-r} dr$$
which is 
$$
2i \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr.$$
By the Cauchy residue theorem, the integral over the contour is
$$
-2\pi i \textrm{Res}_{z=\infty} f(z) = 2\pi i \textrm{Res}_{z=0} \frac{1}{z^2} f(\frac 1 z).$$
From a long and tedious calculation of residue, it turns out that the value on the right is 
$$
2i \frac{\pi e}{24}.$$
Then we have the result:
$$ 
\int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$
