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I'm trying to prove that for $p,q>0$, we have $$\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$ The hint given suggest that we express $\Gamma(p)\Gamma(q)$ as a double integral, then do a change of variables, but I've been unable thus far to express it as a double integral.

Can anyone get me started or suggest an alternate approach?

Note: This wasn't actually given to me as the $\Gamma$ function, just as a function $f$ satisfying $$f(p)=\int_0^\infty e^{-t}t^{p-1}\,dt$$ for all $p>0$, but I recognized that. This is in the context of an advanced calculus practice exam.

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Take a look here this intergal is called beta function, en.wikipedia.org/wiki/Beta_function –  clark Jun 29 '12 at 22:46
    
Similar: math.stackexchange.com/questions/34740/… –  Argon Jun 29 '12 at 23:41

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up vote 8 down vote accepted

Indeed, one can start from the double integral $$ \Gamma(p)\cdot\Gamma(q)=\int_0^\infty \mathrm e^{-x}x^{p-1}\,\mathrm dx\cdot\int_0^\infty \mathrm e^{-y}y^{q-1}\,\mathrm dy=\iint_{[0,+\infty)^2} \mathrm e^{-x-y}x^{p-1}y^{q-1}\,\mathrm dx\,\mathrm dy. $$ The change of variable $x=ts$, $y=(1-t)s$, with $0\leqslant t\leqslant1$ and $s\geqslant0$, whose Jacobian is $\mathrm dx\,\mathrm dy=s\,\mathrm ds\,\mathrm dt$, yields $$ \Gamma(p)\cdot\Gamma(q)=\int_0^{+\infty}\int_0^1\mathrm e^{-s}t^{p-1}s^{p-1}(1-t)^{q-1}s^{q-1}s\,\mathrm ds\,\mathrm dt. $$ Finally, $$\Gamma(p)\cdot\Gamma(q)=\int_0^{+\infty}\mathrm e^{-s}s^{p+q-1}\,\mathrm ds\cdot\int_0^1t^{p-1}(1-t)^{q-1}\,\mathrm dt=\Gamma(p+q)\cdot\int_0^1t^{p-1}(1-t)^{q-1}\,\mathrm dt, $$ and this is it. The last integral above is called the beta number $\mathrm B(p,q)$.

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Clean, pretty, nice. –  ncmathsadist Jun 29 '12 at 22:32
    
@ncmathsadist Thanks. –  Did Jun 30 '12 at 6:14
    
Lovely! And now I feel silly for not choosing a different dummy variable for one of them so I could combine the product as a double integral. –  Cameron Buie Jun 30 '12 at 19:21

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