# Calculating $\int_{\mathcal{S}}x_1^r \, \mathrm dx_1\ldots \, \mathrm dx_n$ [closed]

I need help with the calculation of the following integral

$$\int_{\mathcal{S}}x_1^r \, \mathrm dx_1\ldots \, \mathrm dx_n$$ where $r>0$ and $$\mathcal{S} = \left\{(x_1,\ldots,x_n):a-\epsilon\leq x_1+\ldots+x_n\leq a,\;x_1\ldots,x_n\geq0\right\}$$ for $a>0$ and $a-\epsilon>0$.

Thank you

## closed as off-topic by Carl Mummert, Xam, A. Goodier, Saad, Xander HendersonMar 24 '18 at 3:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Xam, A. Goodier, Saad, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

• Is $r$ an integer (otherwise you have to define $x_1^r$ for negative $x_1$)? Or maybe $\epsilon < a$? – Fabian Dec 23 '12 at 12:15
• @Fabian Right. See edit. – user39097 Dec 23 '12 at 12:18
• Do you also have $x_i>0$ ? – Eckhard Dec 23 '12 at 12:23
• @Eckhard Yes I have. – user39097 Dec 23 '12 at 12:27

We have $$I = \int_{\mathcal{S}}x_1^r\,dx_1\ldots dx_n = \int_{-\infty}^\infty d\lambda \int_{\mathcal{S}} \delta(\lambda-\sum_i x_i) x_1^r\,d^n x = \int_{a-\epsilon}^a d\lambda \int_{x_i\ge 0}\delta(\lambda-\sum_i x_i) x_1^r\,d^nx.$$
Using the integral representation of the $\delta$-function (see Laplace transform) $$\delta(\lambda-\sum_i x_i) = \int_{c-i\infty}^{c+i\infty} \frac{ds}{2\pi i} e^{s(\lambda-\sum_i x_i)}$$ yields \begin{align}I&= \int_{c-i\infty}^{c+i\infty}\frac{ds}{2\pi i} \int_{a-\epsilon}^a d\lambda\int_{x_i\ge 0} e^{s(\lambda-\sum_i x_i)} x_1^r\,d^nx\\ &= \int_{c-i\infty}^{c+i\infty}\frac{ds}{2\pi i} \int_{a-\epsilon}^a d\lambda e^{s\lambda} \int_0^\infty dx_1 x_1^r e^{-s x_1} \left(\int_0^\infty dx e^{-s x}\right)^{n-1}\\ &=\int_{c-i\infty}^{c+i\infty}\frac{ds}{2\pi i} \frac{e^{a s}-e^{s (a-\epsilon )}}{s} \frac{\Gamma(1+r)}{s^{r+1}} \frac{1}{s^{n-1}}\\ &= \frac{\Gamma (r+1) \left(a^{n+r}-(a-\epsilon )^{n+r}\right)}{\Gamma (n+r+1)}. \end{align}
• I tried to implement your "algorithm" to another "similar" integral. However, i'm afraid that it stuck with non-analytically integrals. Specifically, I want to calculate $\int_S(1-x_1^2)^rdx_1,...,dx_n$ where $S = (x_1,...,x_n):a-\epsilon<x_1+...+x_n<a, 1-x_i^2>0$. Do you have any idea? – user39097 Dec 23 '12 at 13:59
• @user39097: not directly. I'm not sure if I understand all the constraints (are the $0<x_i<1$ or $|x_i|<1$?) In principle you can ask another question. Maybe with a reference to this question... – Fabian Dec 23 '12 at 14:04
• I meant $|x_i|<1$. You right. I will open a new post. Thanks – user39097 Dec 23 '12 at 14:09