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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.

Community
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Fedor Petrov
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  • Isn't what Pochhammer formula is about? – Loïc Teyssier Jan 16 '16 at 14:03
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    Pochhammer formula relates integral over $[0,1]$ and integral over Pochhammer contour, but how does it express any of above integrals via $\Gamma$-function? – Fedor Petrov Jan 16 '16 at 14:09
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    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 http://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1/25973#25973. – Pietro Majer Jan 16 '16 at 18:22
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    @Pietro it is probably most common proof, what I call "Fubini". – Fedor Petrov Jan 16 '16 at 19:27
  • ok; $ \int u*v=(\int u)(\int v)$ is indeed Fubini – Pietro Majer Jan 16 '16 at 20:59
  • Real integrals such as those for the gamma function and Riemann zeta (standard Mellin transforms for Re(s) > 0) can be expanded to complex integrals using the Hankel contour, just as Riemann did in his famous paper on the Reimann zeta. The same can be done for the beta integral, which is a particularly simple Mellin transform at the foundations of one class of fractional calculus that has been developed any number of ways and for which Euler developed his iconic integral for the gamma function. Which contour integral are you really looking for? – Tom Copeland Jan 20 '16 at 18:41
  • @TomCopeland say, parallelogram here is perfect: http://mathoverflow.net/questions/105457/complex-evaluation-of-a-classical-real-integral – Fedor Petrov Jan 20 '16 at 18:53
  • We may take an integral using a technique based on Slater's Theorem (cf. O.I.Marichev's book). In fact it will use in an inderect way contour integrals and residues. – Sergei Feb 06 '16 at 16:24

3 Answers3

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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.

Loïc Teyssier
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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.

Kostya_I
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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).

Sergei
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  • This proof is the most popular, I think. It uses change of order in double integral, i.e. some variant of Fubini theorem. – Fedor Petrov Jan 16 '16 at 14:58