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?

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.
