How to compute the integral $\int_{0}^{\infty}...\int_{0}^{\infty} \frac{1}{(2+\sum_{i=1}^{5}x_i)^6} dx_1...dx_5$?

Question

How to compute the integral $\int_{0}^{\infty}...\int_{0}^{\infty} \frac{1}{(2+\sum_{i=1}^{5}x_i)^6} dx_1...dx_5$?

In general, these kind of integrals represent an integral of a probability density function over $[0, \infty)^5$. But I cannot think of any standard density of this form. Is there any other way to approach this?

Answer 1

Well, we are trying to find:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right):=\int\limits_0^\infty\cdots\underbrace{\int\limits_0^\infty\frac{1}{\displaystyle\left(\alpha+\sum_{\text{m}\space=\space1}^\text{n}x_\text{m}\right)^\text{k}}\space\text{d}x_1}_{:=\space\mathcal{S}_1}\cdots\space\text{d}x_\text{n}\tag1$$

Now, we can see that:

$$\mathcal{S}_1=\int\limits_0^\infty\frac{1}{\displaystyle\left(\alpha+\sum_{\text{m}\space=\space1}^\text{n}x_\text{m}\right)^\text{k}}\space\text{d}x_1\tag2$$

So, let's write:

$$\mathcal{A}_1=\alpha+\sum_{\text{m}\space=\space1}^\text{n}x_\text{m}=x_1+\underbrace{\alpha+\sum_{\text{m}\space=\space2}^\text{n}x_\text{m}}_{:=\space\beta_2}\tag3$$

So, we want to find:

$$\mathcal{S}_1=\int\limits_0^\infty\frac{1}{\displaystyle\left(x_1+\beta_2\right)^\text{k}}\space\text{d}x_1=\frac{\beta_2^{1-\text{k}}}{\text{k}-1}=\frac{\displaystyle\left(\alpha+\sum_{\text{m}\space=\space2}^\text{n}x_\text{m}\right)^{1-\text{k}}}{\text{k}-1}\tag4$$

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So, for your case we get:

\begin{equation} \begin{split} \mathcal{I}_5\left(6,2\right)&=\int\limits_0^\infty\cdots\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space1}^5x_\text{m}\right)^6}\space\text{d}x_1\cdots\space\text{d}x_5\\ \\ &=\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space1}^5x_\text{m}\right)^6}\space\text{d}x_1\space\text{d}x_2\space\text{d}x_3\space\text{d}x_4\space\text{d}x_5 \end{split}\tag5 \end{equation}

Using $(4)$ we can see:

$$\mathcal{S}_1=\int\limits_0^\infty\frac{1}{\displaystyle\left(x_1+\beta_2\right)^6}\space\text{d}x_1=\frac{\displaystyle\left(2+\sum_{\text{m}\space=\space2}^\text{n}x_\text{m}\right)^{1-6}}{6-1}=\frac{1}{5}\cdot\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space2}^\text{n}x_\text{m}\right)^5}\tag6$$

So, we get (which is not hard to see I think):

\begin{equation} \begin{split} \mathcal{I}_5\left(6,2\right)&=\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\frac{1}{5}\cdot\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space2}^5x_\text{m}\right)^5}\space\text{d}x_2\space\text{d}x_3\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space2}^5x_\text{m}\right)^5}\space\text{d}x_2\space\text{d}x_3\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\frac{1}{4}\cdot\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space3}^5x_\text{m}\right)^4}\space\text{d}x_3\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\int\limits_0^\infty\int\limits_0^\infty\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space3}^5x_\text{m}\right)^4}\space\text{d}x_3\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\int\limits_0^\infty\int\limits_0^\infty\frac{1}{3}\cdot\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space4}^5x_\text{m}\right)^3}\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\int\limits_0^\infty\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space4}^5x_\text{m}\right)^3}\space\text{d}x_4\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\int\limits_0^\infty\frac{1}{2}\cdot\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space5}^5x_\text{m}\right)^2}\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\int\limits_0^\infty\frac{1}{\displaystyle\left(2+\sum_{\text{m}\space=\space5}^5x_\text{m}\right)^2}\space\text{d}x_5\\ \\ &=\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{1}\cdot\frac{1}{\displaystyle\left(2+0\right)^1}\\ \\ &=\frac{1}{240} \end{split}\tag7 \end{equation}

Now, we can prove that:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\frac{1}{\left(\text{k}-1\right)!}\cdot\frac{1}{\alpha^{\text{k}-\text{n}}}\tag8$$

Answer 2

Hint

Consider the most general integral (with $a >0$) $$\int_{0}^{\infty}\cdots\int_{0}^{\infty} \frac{1}{(a+\sum_{i=1}^{n}x_i)^{n+1}}\,dx_1\cdots dx_n $$

For the most inner integral, define $$A_1=a+\sum_{i=2}^{n}x_i\implies \int_{0}^{\infty} \frac{1}{(a+\sum_{i=1}^{n}x_i)^{n+1}}\,dx_1=\int_{0}^{\infty} \frac{dx_1}{(A_1+x_1)^{n+1}}=\frac 1 n\frac 1{A_1^n}$$ Replace $A_1$ by its definition and repeat the process till the end.

As @mjqxxxx commented, this is more than simple.