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?
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: $$ \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$
Torsten Carleman $[2]$ proved in 1922 that
$$ \sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty a_n, $$
where $a_n \geq 0$, $n=1,2,\dots$, and $0 < \sum_{n=1}^\infty a_n < \infty$. Thenceforth, this result is known as Carleman's inequality. There exists a number of refined versions of Carleman's original work $[3, 6]$. It has turned out that the following generalization is – from our point of view – important, which is proved by Yang $[7]$: $$ \sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty \left(1-\sum_{k=1}^6 \frac{b_k}{(n+1)^k}\right)a_n, $$
with $b_1 = 1/2, b_2 = 1/24, b_3 = 1/48, b_4 = 73/5670, b_5 = 11/1280, b_6 = 1945/580608$. On the last page of his paper, Yang $[7]$ conjectured that if $$ \left(1+\frac{1}{x}\right)^x = e\left(1-\sum_{n=1}^\infty \frac{b_n}{(x+1)^n}\right), \quad x>0, $$
then $b_n > 0$, $n=1,2,\dots.$ In fact, the constants $b_4$ and $b_6$ are not corrent in Yang's work, the correct values are $b_4 = 73/5760$ and $b_6 = 3625/580608$. Later, this conjecture was proved and discussed by Yang $[8]$, Gylletberg and Ping $[4]$, and Yue $[9]$. They are using the recurrence $$ b_1 = \frac12, \quad b_n = \frac{1}{n}\left(\frac{1}{n+1} - \sum_{k=0}^{n-2} \frac{b_{n-k-1}}{k+2} \right), \quad n = 2,3,\dots. $$ The recurrence is given in a somewhat more compact form in Finch's manuscript $[3]$, as the following:
$$ b_0 = -1, \quad b_n = -\frac{1}{n}\sum_{k=1}^{n} \frac{b_{n-k}}{k+1}, \quad n = 1,2,\dots. $$
The first ten values of the sequence are listed in the next table. \begin{array} {|r|r|} \hline b_0 & -1 \\ \hline b_1 & 1/2 \\ \hline b_2 & 1/24 \\ \hline b_3 & 1/48 \\ \hline b_4 & 73/5760 \\ \hline b_5 & 11/1280 \\ \hline b_6 & 3625/580608 \\ \hline b_7 & 5525/1161216 \\ \hline b_8 & 5233001/1393459200 \\ \hline b_9 & 1212281/398131200 \\ \hline b_{10} & 927777937/367873228800 \\ \hline \end{array} The numerators are recorded as A249276, and the denominators as A249277 in the OEIS. I've calculated the $b_n$ sequence in the range $n=0,\dots,32$, the elements are listed here.
The following theorem is proved in general in the paper by Hu and Mortici $[5]$, and for the special cases $n=0$ and $n=1$ in the paper by Alzer and Berg $[1]$.
For all integer $n \geq 0$, we have
$$ \int_0^1 x^n\sin\left(\pi x\right)x^x\left(1-x\right)^{1-x}\,dx = b_{n+2}\pi e. $$
The special case $n=0$ answers my question.
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