Two curious "identities" on $x^x$, $e$, and $\pi$ A numerical calculation on Mathematica shows that
$$I_1=\int_0^1 x^x(1-x)^{1-x}\sin\pi x\,\mathrm dx\approx0.355822$$
and
$$I_2=\int_0^1 x^{-x}(1-x)^{x-1}\sin\pi x\,\mathrm dx\approx1.15573$$
A furthur investigation on OEIS (A019632 and A061382) suggests that $I_1=\frac{\pi e}{24}$ and $I_2=\frac\pi e$ (i.e., $\left\vert I_1-\frac{\pi e}{24}\right\vert<10^{-100}$ and $\left\vert I_2-\frac\pi e\right\vert<10^{-100}$).
I think it is very possible that $I_1=\frac{\pi e}{24}$ and $I_2=\frac\pi e$, but I cannot figure them out. Is there any possible way to prove these identities?
 A: Supplementary calculation of residue of the function
$$f(z)=e^{z\frac{e^z}{e^z-1}}\frac{e^z}{(e^z-1)^3}$$
at its triple pole $z=0$:
$f(z)$ is an odd function.Then 
$$g(z)=f(z)^{\frac{1}{3}}=e^{\frac{1}{3}z(\frac{e^z}{e^z-1}+1)}\frac{1}{e^z-1}$$
is an odd function as well.
Hence 
$$g(z)=\frac{A_0}{z}+A_1 z+\cdots$$
and
$$Resf(z)_{\vert z=0}=3A_0^2A_1.$$
$z=0$ is a simple pole,then we can get $A_0=e^{\frac{1}{3}}$ without hesitation.
$$\frac{1}{e^z-1}=\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+\cdots=\frac{a_0}{z}+a_1+a_2 z+\cdots$$
$$\frac{z}{1-e^{-z}}+z=1+\frac{3}{2}z+\frac{1}{12}z^2+\cdots=b_0+b_1z+b_2 z^2+\cdots$$
Hence
$$\exp{(b_0+b_1z+b_2 z^2+\cdots)}=\exp(b_0)+b_1\exp(b_0)z+(b_1^2/2+b_2)\exp(b_0)z^2+\cdots,$$
i.e.,
$$e^{\frac{1}{3}z(\frac{e^z}{e^z-1}+1)}=e^{\frac{1}{3}}+\frac{1}{2}e^{\frac{1}{3}}z+\frac{11}{72}e^{\frac{1}{3}}z^2+\cdots$$
And $$A_1=\frac{1}{12}e^{\frac{1}{3}}-\frac{1}{2}\frac{1}{2}e^{\frac{1}{3}}+\frac{11}{72}e^{\frac{1}{3}}=-\frac{1}{72}e^{\frac{1}{3}}$$
Therefore $Resf(z)_{\vert z=0}=3A_0^2A_1=-e/24$.
A: You made a very nice observation! Often it is important to make a good guess than just to solve a prescribed problem. So it is surprising that you made a correct guess, especially considering the complexity of the formula.
I found a solution to the second integral in here, and you can also find a solution to the first integral at the link of this site.
A: Somebody actually made a very interesting tactic on MathOverflow. Here is a link to that idea.
It mainly involved these two integral representations and multiplying the two (Euler Reflection Formula). 

Indeed, letting $\ell(u):=u-\ln u$ for $u>0$, note that for $x\in(0,1)$ 
  $$\qquad \Gamma(x)x^{-x}=\int_0^\infty e^{-x\ell(u)}\,du=\int_0^\infty e^{-x\ell(u)}\,\frac{du}u
$$ 
  and also
  $$\qquad \Gamma(1-x)(1-x)^{x-1}=\int_0^\infty e^{(x-1)\ell(v)}\,dv=\int_0^\infty e^{(x-1)\ell(v)}\,\frac{dv}v.  
$$

This is really for the sake of alternate ideas.
