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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 ?.

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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

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

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  • $\begingroup$ (+1) This proves my alternative computation was right. $\endgroup$ – Jack D'Aurizio Jan 6 '15 at 20:52
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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: $$ \color{red}{I=\frac{\Gamma(p+1)\Gamma(q+1)\Gamma(r+1)\Gamma(s+1)}{\Gamma(4+p+q+r+s)}}. $$

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  • $\begingroup$ Well, thanks. My first question is what do you mean by $d\mu$? Do you mean $d\mu(u,v,w)$? And my second question is there an alternative solution using the change I wrote above or this one is just simpler? $\endgroup$ – nilcorc Jan 6 '15 at 20:31
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    $\begingroup$ @Novsar: yes, $d\mu$ is a shortcut notation for $du\,dv\,dw$ in the first case and for $d\theta\,d\phi$ in the second case. I haven't tried your change of variables since this one looked to me as the most natural way for separating variables and "factor" the integral, but probably your way is as efficient as mine. $\endgroup$ – Jack D'Aurizio Jan 6 '15 at 20:48
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    $\begingroup$ @Novsar: Ok, the achille hui's answer confirms the feeling. $\endgroup$ – Jack D'Aurizio Jan 6 '15 at 20:51
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    $\begingroup$ @Novsar: it is due to some change of variable bringing such integrals in the usual form, suited for beta-function evaluation. For instance: $$\int_{0}^{\pi/2}\sin^\alpha(\theta)\,d\theta = \int_{0}^{1} t^{\alpha}(1-t^2)^{-1/2}\,dt = \color{red}{\frac{1}{2}}\int_{0}^{1}u^{(\alpha-1)/2}(1-u)^{-1/2}\,du.$$ $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 20:19
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    $\begingroup$ I've just found out why. Thank you a lot for help. I really liked your solution to the problem. :) $\endgroup$ – nilcorc Jan 8 '15 at 20:23
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{V}\ x^{p}\,y^{q}\,z^{r}\ \pars{1 - x - y - z}^{\,s}\,\dd x\,\dd y\,\dd z\,;\ p, q, r, s >0}$ and $\ds{V=\braces{\pars{ x,y,z} \in {\mathbb R}^{3}_{+}: x + y + z\ \leq\ 1}}$


\begin{align}&\color{#66f}{\large\left. \int_{V}\ x^{p}\,y^{q}\,z^{r}\pars{1 - x - y - z}^{\, s}\,\dd x\,\dd y\,\dd z\, \right\vert_{\, x + y + z\ < 1}} \\[5mm]&=\left.\int_{0}^{\infty}\int_{0}^{\infty} \int_{0}^{\infty}x^{p}\,y^{q}\,z^{r} \pars{1 - x - y - z}^{\,s}\,\dd x\,\dd y\,\dd z\, \right\vert_{\, x + y + z\ < 1} \\[5mm]&=\int_{0}^{\infty}\int_{0}^{\infty} \int_{0}^{\infty}x^{p}\,y^{q}\,z^{r}\ \Theta\pars{1 - x - y - z}\pars{1 - x - y - z}^{\,s}\,\dd x\,\dd y\,\dd z \\[5mm]&=\int_{0}^{\infty}\int_{0}^{\infty} \int_{0}^{\infty}x^{p}\,y^{q}\,z^{r}\ \overbrace{\int_{0^{-}}^{\infty} \delta\pars{1 - x - y - z - \xi}\xi^{\,s}\,\dd\xi} ^{\dsc{\Theta\pars{1 - x - y - z}\pars{1 - x - y - z}^{\,s}}}\ \,\dd x\,\dd y\,\dd z \\[5mm]&=\int_{0}^{\infty}\int_{0}^{\infty} \int_{0}^{\infty}\int_{0^{-}}^{\infty}x^{p}\,y^{q}\,z^{r}\xi^{\,s}\ \overbrace{% \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} \exp\pars{\tau\bracks{1 - x - y - z - \xi}}\,{\dd\tau \over 2\pi\ic}} ^{\dsc{\delta\pars{1 - x - y - z - \xi}}}\,\,\,\, \,\dd x\,\dd y\,\dd z\,\dd\xi \\[5mm]&=\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\ \overbrace{\int_{0}^{\infty}x^{p}\expo{-\tau x}\,\dd x} ^{\dsc{\tau^{-p - 1}\ \Gamma\pars{p + 1}}}\ \underbrace{\int_{0}^{\infty}y^{q}\expo{-\tau y}\,\dd y} _{\dsc{\tau^{-q - 1}\ \Gamma\pars{q + 1}}}\ \overbrace{\int_{0}^{\infty}z^{r}\expo{-\tau z}\,\dd z} ^{\dsc{\tau^{-r - 1}\ \Gamma\pars{r + 1}}}\ \underbrace{\int_{0}^{\infty}\xi^{s}\expo{-\tau\xi}\,\dd\xi} _{\dsc{\tau^{-s - 1}\ \Gamma\pars{s + 1}}}\,\,\,\, \expo{\tau}\,{\dd\tau \over 2\pi\ic} \\[5mm]&=\Gamma\pars{p + 1}\Gamma\pars{q + 1}\Gamma\pars{r + 1}\Gamma\pars{s + 1}\ \underbrace{\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} \frac{\expo{\tau}}{\tau^{p + q + r + s + 4}}\,\,\,{\dd\tau \over 2\pi\ic}} _{\ds{\dsc{1 \over \pars{p + q + r + s + 3}!}\ =\ \dsc{1 \over \Gamma\pars{p + q + r + s + 4}}}} \end{align}
Finally, \begin{align} &\color{#66f}{\large\left.% \int_{V}\ x^{p}\,y^{q}\,z^{r}\pars{1 - x - y - z}^{\,s}\,\dd x\,\dd y\,\dd z \,\right\vert_{\, x + y + z\ <\ 1}} \\[5mm]&=\color{#66f}{\large{\Gamma\pars{p + 1}\Gamma\pars{q + 1}\Gamma\pars{r + 1}\Gamma\pars{s + 1} \over \Gamma\pars{p + q + r + s + 4}}} \end{align}

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