It is a question in spirit of this one. Is there a way to prove Euler's formula $$ \int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $$ using contour integration (and maybe something else, say, integration by parts or change of variables is ok, but double integration and Fubini theorem is not)? For $a=b=1/2$ we get the value of $\Gamma(1/2)$ which is essentially the value the cited question was about.
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asked Jan 16, 2016 at 9:41
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$\begingroup$ Isn't what Pochhammer formula is about? $\endgroup$– Loïc TeyssierJan 16, 2016 at 14:03
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1$\begingroup$ Pochhammer formula relates integral over $[0,1]$ and integral over Pochhammer contour, but how does it express any of above integrals via $\Gamma$-function? $\endgroup$– Fedor PetrovJan 16, 2016 at 14:09
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1$\begingroup$ The way I like to of see that identity is, seeing it as a consequence of the fact that the integral of a convolution is the product of integrals, as mentioned here mathoverflow.net/questions/20960/…. $\endgroup$– Pietro MajerJan 16, 2016 at 18:22
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2$\begingroup$ @Pietro it is probably most common proof, what I call "Fubini". $\endgroup$– Fedor PetrovJan 16, 2016 at 19:27
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$\begingroup$ ok; $ \int u*v=(\int u)(\int v)$ is indeed Fubini $\endgroup$– Pietro MajerJan 16, 2016 at 20:59
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3 Answers
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In what follows we assume $\Re(a)>0$ and $\Re(b)>0$.
Begin with the case $a+b=k\in\mathbb N$. Using Pochhammer contour $P$, one can relate what's going on on $[0,1]$ to what is going on on a circle $C:=x_*\mathbb S^1$, $|x_*|>1$. Indeed, looking carefully at determinations of $f(z):=z^{a-1}(1-z)^{b-1}$ one has $$ \oint_Pf(z)dz = (1-\exp 2ib\pi)\oint_C f(z)dz ~~~(*)\\=(1-\exp 2ib\pi)(1-\exp 2ia\pi)\int_0^1f(z)dz$$ the last equality being given by Pochhammer formula, so that $$ \oint_C f(z)dz = 2i(-1)^{1-a}\sin(a\pi) \int_0^1f(z)dz .$$ Since $f(z)=z^{a+b-2}(1/z-1)^{b-1}$ is holomorphic near $\infty$ we have $$\oint_Cf(z)dz = -\oint_C f(1/x)\frac{dx}{x^2}=(-1)^{b-1}\oint_C x^{-k}(1-x)^{b-1}dx , $$ which is a contour integral. It can be evaluated by looking a the expansion of $$(1-x)^{b-1} = \sum_n \frac{\Gamma(b)}{\Gamma(n+1)\Gamma(b-n)}(-x)^n .$$ The residue of $x^{-k}(1-x)^{b-1}$ at $0$ is obtained for $n+1=k$, that is $\frac{\Gamma(b)}{\Gamma(a+b)\Gamma(1-a)}$ which allows to conclude using Gamma reflection formula $\Gamma(1-a)\Gamma(a)\sin(a\pi)=\pi$.
The next step is to deal with the case $a+b=p/q\in\mathbb Q$, then conclude by analyticity and accumulation. This case is dealt with by taking a linear combination of $\oint_Cf(z)dz$ with weights $\exp (2in\pi/q)$ to obtain the same kind of relation as $(*)$. I'll write details later, but they should be straightforward.
answered Jan 16, 2016 at 22:35
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$\begingroup$ Gamma reflection formula is a partial case and I do not see immediately independent proof of it. $\endgroup$ Jan 16, 2016 at 22:49
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There is a proof in R. Remmert's "Classical topics in complex function theory", which is pretty complex analytic. It uses a uniqueness theorem for functions satisfying $v(z+1)=zv(z)$ (derived from Liouville), and functional equations for $\Gamma$ and $B$ functions (derived by integration by parts). No double integrals or Fubini, but, unfortunately, no explicit contour integrals either.
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$\begingroup$ Well, not bad, but contour integral would be better) $\endgroup$ Jan 16, 2016 at 18:02
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There is a proof using only change of variables in N.N.Lebedev's book (p.28 of the Russian edition).
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$\begingroup$ This proof is the most popular, I think. It uses change of order in double integral, i.e. some variant of Fubini theorem. $\endgroup$ Jan 16, 2016 at 14:58