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How one can show this equality? $$ \iiint_V x^{a-1}y^{b-1}z^{c-1}\,dxdydz = \dfrac{\Gamma(a)\Gamma(b)\Gamma(c)}{\Gamma(a+b+c+1)}, $$ where $V$ is simplex $x\geqslant0, y\geqslant0, z\geqslant0, x+y+z\leqslant1 $.

Only thing I thought is changing variable in some way but it didn't help me.

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I'm thinking about changing variable as $u=-\ln x, v =-\ln y , w=-\ln z$ – Nikita Evseev Apr 5 '13 at 8:44

That's actually quite a nice one. If you know Quantum Field Theory you might have heard of Feynman parameters? They come in handy here (beware rigour, I'm a physicist :D) So, how can you prove the above? First note

$$\frac{1}{AB}=\int_{0}^1dxdy \delta(x+y-1)\frac{1}{(xA+yB)^2}.$$ This can be proven straightforwardly. Moreover, it can be generalized

$$\frac{1}{A_1\cdots A_n}=\int_0^1 dx_1 \cdots dx_n \delta\left(\sum_i x_i -1\right)\frac{(n-1)!}{(x_1A_1+\ldots+x_nA_n)^n}$$ (Proof by induction.)

Now, you can differentiate wrt to the $A_i$ to find

$$\frac{1}{A_1^{m_1} \cdots A_n^{m_n}}=\int_0^1 dx_1 \cdots dx_n \delta\left(\sum_i x_i-1\right)\frac{\prod_i x_i^{m_i-1}}{\left(\sum_i x_i A_i\right)^{\sum m_i}}\cdot \frac{\Gamma(m_1+\ldots+m_n)}{\Gamma(m_1)\Gamma(m_2) \cdots \Gamma(m_n)}.$$

Set all $A_i$ to one and you essentially have your statement. This is also stated in the QFT book by Peskin/Schroeder on p. 190.

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Well I'm going to spend some time to understand this (actually now I'm taking a course in QM). Unfortunately I should explain how to solve this problem to my calculus class and I think that my students have not yet heard about QFT. So I looking for some of elementary explanation. – Nikita Evseev Apr 5 '13 at 9:28
Well, there's actually no reference in my derivation to QFT. Just some playing around with integral identities. I just meant that these sorts of integrals appear in QFT loop computations...that's all :) Other than that, it's pretty straightforward maths :) – A friendly helper Apr 5 '13 at 9:33
This is very nice even though the method doesn't work for non-integer $a,b,c$. (+1) – Start wearing purple May 20 '13 at 15:09

Here is how the solution,

$$ I=\iiint_V x^{a-1}y^{b-1}z^{c-1}\,dx\,dy\,dz = \int_{0}^{1}x^{a-1}dx\int_{0}^{1-x}y^{b-1}dy\int_{0}^{1-x-y}z^{c-1}dz $$


$$I = \frac{1}{c}\int_{0}^{1}x^{a-1}dx\int_{0}^{1-x}y^{b-1}(1-x-y)^c\,dy $$

Using the change of variables $t=\frac{y}{1-x}$ and beta function after factoring out $(1-x)$, we have

$$ I = \frac{1}{c}{\frac{\Gamma\left( c \right) \Gamma \left( b \right) }{\Gamma\left(c+b+1 \right) }} \int_{0}^{1}x^{a-1}(1-x)^{c+b}\,dx $$

$$ I = {\frac{\Gamma\left( c \right) \Gamma \left( b \right) }{\Gamma\left(c+b+1 \right) }}\frac{\Gamma(a)\Gamma(c+b+1)}{\Gamma(a+b+c+1)} = \frac{\Gamma(a)\Gamma(b)\Gamma(c)}{\Gamma(a+b+c+1)}.$$

Note: The beta function is defined as

$$ \beta(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt \!,\quad \textrm{Re}(x), \textrm{Re}(y) > 0.\, $$

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You remember beta-function? $B(z,w)=\displaystyle \int_0^1 t^{z-1}(1-t)^{w-1} dt$, you are going to use it and make some substitutions. You take the integral one at a time while making the substitution and see what happen. Am sure you will get it.

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I am sure the Riemann Hypothesis is correct. Yet these mathematicians insist on trying to prove it. – Ron Gordon Apr 5 '13 at 9:05
Ron Gordon what is your point? – YYG Apr 5 '13 at 9:15
You failed to present an answer. – Ron Gordon Apr 5 '13 at 9:16
@RonGordon, sorry for my lack of knowledge on the subject, but what Riemann's Hypothesis has to do with YYG's answer? – user1620696 May 4 '13 at 2:54
@user1620696: nothing. Zero. Which is about as much use as this answer is in solving the problem. – Ron Gordon May 4 '13 at 5:28

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