Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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. –  froggie Nov 22 '12 at 14:33
$10^{-100}$ 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/… –  Alexander Gruber Nov 22 '12 at 16:16
How about this? –  sos440 Nov 24 '12 at 14:05
@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... –  Gottfried Helms Nov 24 '12 at 14:10
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/… –  Gottfried Helms Nov 24 '12 at 14:14
show 6 more comments

2 Answers

up vote 16 down vote accepted

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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