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

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

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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