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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?

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    $\begingroup$ 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
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    $\begingroup$ $10^{-100}$ 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
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    $\begingroup$ How about this? $\endgroup$ – Sangchul Lee Nov 24 '12 at 14:05
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    $\begingroup$ @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... $\endgroup$ – Gottfried Helms Nov 24 '12 at 14:10
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    $\begingroup$ 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/… $\endgroup$ – Gottfried Helms Nov 24 '12 at 14:14
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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.

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

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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.

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

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