A numerical calculation on Mathematica shows that

$$I_1=\int_0^1 x^x(1-x)^{1-x}\sin\pi x\,\mathrm dx\approx0.355822$$


$$I_2=\int_0^1 x^{-x}(1-x)^{x-1}\sin\pi x\,\mathrm dx\approx1.15573$$

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?

  • 7
    $\begingroup$ Using the Euler Reflection Formula $\sin\pi x = \pi/\Gamma(x)\Gamma(1-x)$ one can rewrite $I_1$ as $$I_1 = \pi\int_0^1 \frac{x^x(1-x)^{1-x}}{\Gamma(x)\Gamma(1-x)}\,dx.$$ Your "identities," if true, are then really just statements about $e$ and not about both $e$ and $\pi$. I'm not sure if this simplification helps at all. There is also the identity $\Gamma(x)\Gamma(1-x) = B(x,1-x)$, where $B$ is the Beta function. $\endgroup$
    – froggie
    Nov 22 '12 at 14:33
  • 2
    $\begingroup$ $10^{-100}$ is crazy small. If it turns out these aren't what you think they are, they'd make a fantastic addition to this list: math.stackexchange.com/questions/111440/… $\endgroup$
    – Alexander Gruber
    Nov 22 '12 at 16:16
  • 11
    $\begingroup$ How about this? $\endgroup$ Nov 24 '12 at 14:05
  • 6
    $\begingroup$ @sos440 (+1): why not make this an answer? Before reading your hint I made sure that the identity holds to 800 digits precision using Pari/GP... $\endgroup$ Nov 24 '12 at 14:10
  • 2
    $\begingroup$ In the link which sos440 points to it is a reference into a list of formulae in wikipedia, attributed to Ramanujan, see: de.wikibooks.org/wiki/… $\endgroup$ Nov 24 '12 at 14:14

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.


Supplementary calculation of residue of the function


at its triple pole $z=0$:

$f(z)$ is an odd function.Then


is an odd function as well.

Hence $$g(z)=\frac{A_0}{z}+A_1 z+\cdots$$


$$Resf(z)_{\vert z=0}=3A_0^2A_1.$$

$z=0$ is a simple pole,then we can get $A_0=e^{\frac{1}{3}}$ without hesitation.

$$\frac{1}{e^z-1}=\frac{1}{z}-\frac{1}{2}+\frac{1}{12}z+\cdots=\frac{a_0}{z}+a_1+a_2 z+\cdots$$

$$\frac{z}{1-e^{-z}}+z=1+\frac{3}{2}z+\frac{1}{12}z^2+\cdots=b_0+b_1z+b_2 z^2+\cdots$$


$$\exp{(b_0+b_1z+b_2 z^2+\cdots)}=\exp(b_0)+b_1\exp(b_0)z+(b_1^2/2+b_2)\exp(b_0)z^2+\cdots,$$



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 $Resf(z)_{\vert z=0}=3A_0^2A_1=-e/24$.


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.

  • $\begingroup$ Thank you for the upvotes. I do indeed feel that this post needs mentioning. $\endgroup$
    – user311151
    Apr 8 '16 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.