Integral in $\mathbb R^3$ and $\Gamma$-function 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.
 A: 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.
A: Here is how the solution,
$$ I=\iiint_V x^{a-1}y^{b-1}z^{c-1}\,dx\,dy\,dz = \int_{0}^{1}x^{a-1}dx\int_{0}^{1-x}y^{b-1}dy\int_{0}^{1-x-y}z^{c-1}dz $$
Added:
$$I = \frac{1}{c}\int_{0}^{1}x^{a-1}dx\int_{0}^{1-x}y^{b-1}(1-x-y)^c\,dy $$
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 
$$ \beta(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt \!,\quad \textrm{Re}(x), \textrm{Re}(y) > 0.\, $$
A: 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.
