A numerical calculation on Mathematica shows that

I1=∫10xx(1−x)1−xsinπxdx≈0.355822

and

I2=∫10x−x(1−x)x−1sinπxdx≈1.15573

A furthur investigation on OEIS (A019632 and A061382) suggests that I1=πe24 and I2=πe (i.e., |I1−πe24|<10−100 and |I2−πe|<10−100).

I think it is very possible that I1=πe24 and I2=πe, but I cannot figure them out. Is there any possible way to prove these identities?

**Answer**

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.

**Attribution***Source : Link , Question Author : zy_ , Answer Author : Sangchul Lee*