Prove that ∫10sin(πx)xx(1−x)1−xdx=πe24\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24}

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:
$$\int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$