Dirichlet's integral $\int_{V}\ x^{p}\,y^{q}\,z^{r}\ \left(\, 1 - x - y - z\,\right)^{\,s}\,{\rm d}x\,{\rm d}y\,{\rm d}z$

I found such an exercise:

Calculate the Dirichlet's integral:

$$ \int_{V}\ x^{p}\,y^{q}\,z^{r}\ \left(\, 1 - x - y - z\,\right)^{\,s}\,{\rm d}x\,{\rm d}y\,{\rm d}z \quad\mbox{where}\quad p, q, r, s >0 $$ and $V=\left\{\,\left(\, x,y,z\,\right) \in {\mathbb R}^{3}_{+}: x + y + z\ \leq\ 1\right\}$


I thought that I could put $x + y + z = \alpha$. I got a clue, that it is a correct approach, but I should also put $y + z = \alpha\beta$ and $z=\alpha\beta\gamma$. So:

$z=\alpha\beta\gamma\,,\quad y=\alpha\beta\left(\, 1 - \gamma\,\right)\,,\quad x=\alpha\left(\, 1 - \beta\,\right)$

Should I change $x,y,z$ under the integral sign to $\alpha,\beta,\gamma$ now ?.

Solution 1:

For Type I Dirichlet integrals, one has the formula:

$$\int_{\Delta_n} f\left(\sum_{k=1}^n t_k\right) \prod_{k=1}^n t_k^{\alpha_k-1}\prod_{k=1}^n dt_k = \frac{\prod_{k=1}^n \Gamma(\alpha_k)}{\Gamma(\sum_{k=1}^n\alpha_k)}\int_0^1 f(\tau) \tau^{(\sum_{k=1}^n\alpha_k)-1} d\tau$$

where $$\Delta_n = \bigg\{ (x_1,\ldots,x_n) \in [0,\infty)^n : \sum_{k=1}^n x_k \le 1 \bigg\}$$ is the standard $n$-simplex. For a proof of a very similar formula where $\Delta_n$ is replaced by $[0,\infty)^n$, see this answer. It will show you how to carry out the computation in your original approach.

Apply it to your integral with

$$f(w) = (1-w)^s\quad\text{ and }\quad \begin{cases} \alpha_1 = p + 1\\ \alpha_2 = q + 1\\ \alpha_3 = r + 1 \end{cases}, $$ one find

$$\begin{align} \int_{\Delta_3}(1-x-y-z)^s x^p y^q z^r dxdydz = & \frac{\Gamma(p+1)\Gamma(q+1)\Gamma(r+1)}{\Gamma(p+q+r+3)}\int_0^1 (1-\tau)^s t^{p+q+r+2} d\tau\\ = &\frac{\Gamma(p+1)\Gamma(q+1)\Gamma(r+1)\Gamma(s+1)}{\Gamma(p+q+r+s+4)} \end{align} $$

Solution 2:

Why not to map $(x,y,z)$ into $(u^2,v^2,w^2)$ and integrate over a spherical sector?

With the first change of variables we have: $$ I = 8\iiint_S u^{2p+1} v^{2q+1} w^{2r+1} (1-(u^2+v^2+w^2))^s\,d\mu $$ where $S=\{(u,v,w)\in(0,+\infty)^3: u^2+v^2+w^2\leq 1\}$.

By setting $u=\rho\cos\theta\sin\phi, v=\rho\sin\theta\sin\phi, w=\rho\cos\phi$, we get:

$$ I = 8\int_{0}^{1}\iint_{(0,\pi/2)^2}\rho^{2p+2q+2r+5}(1-\rho^2)^s\cos^{2p+1}\theta\sin^{2q+1}\theta\cos^{2r+1}\phi\sin^{2p+2q+3}\phi\,d\mu\,d\rho,$$ but since, due to the properties of the Euler Beta function: $$ \int_{0}^{1}\rho^{2p+2q+2r+5}(1-\rho^2)^s\,d\rho = \frac{\Gamma(3+p+q+r)\Gamma(1+s)}{2\Gamma(4+p+q+r+s)},$$ $$\int_{0}^{\pi/2}\cos^{2p+1}\theta\sin^{2q+1}\theta\,d\theta = \frac{\Gamma(p+1)\Gamma(q+1)}{2\Gamma(2+p+q)},$$ $$\int_{0}^{\pi/2}\cos^{2r+1}\phi\sin^{2p+2q+3}\phi\,d\phi=\frac{\Gamma(2+p+q)\Gamma(1+r)}{2\Gamma(3+p+q+r)},$$ it follows that: $$ \color{red}{I=\frac{\Gamma(p+1)\Gamma(q+1)\Gamma(r+1)\Gamma(s+1)}{\Gamma(4+p+q+r+s)}}. $$