Can contour integration be used to integrate $~u^u(1-u)^{1-u}~$?

Asked Aug 30 '19 at 17:10

Active Aug 30 '19 at 17:16

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I am told contour integration can be used to show

$$\int_0^1 e^{i \pi z} z^z(1-z)^{1-z}=\frac{i\pi e}{24}$$

but am skeptical as the integrand contains log functions which do not admit simple closed contours for example to use the Residue Theorem. However I suspect perhaps a pochhammer contour might work, i.e, a (closed) analytically-continuous figure-8 contour going around $~0~$ and $~1~$ similar to the integral representation of the beta function. Might someone be familiar with this integral and the method used to integrate it and give me some ideas?

Thanks, Dominic

edited Aug 30 '19 at 17:16

nmasanta

9,222

asked Aug 30 '19 at 17:10

Dominic

It looks like Beta Function... – MafPrivate Aug 30 '19 at 17:12
Yes however in the beta function the exponents are held constant. – Dominic Aug 30 '19 at 17:20
The addendum in this old answer of mine appears to be related. – Jack D'Aurizio Aug 30 '19 at 18:23
@metamorphy: Thanks. I'll work through it. – Dominic Aug 30 '19 at 18:35

Can contour integration be used to integrate $~u^u(1-u)^{1-u}~$?

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