How can we change to polar coordiantes and evalute this integral:
$$ \int_{0}^\infty \int_{0}^\infty \left(x+y\right) ^{k} e^{-c(x+y)}d{x}d{y}$$
where $c$ is a positive constant, $k$ is a real positive integer.
Now, can we extend our result to the general case given by
$$ \int_{X_{\ell}} \left(\sum_{\ell=1}^{m}x_{\ell}\right) ^{k} e^{-c \sum_{\ell=1}^{m}x_{\ell}} \prod_{\ell=1}^{m }d{x_{ \ell}} $$
The integration is over $0<{x_{ \ell}}<\infty$ for $\ell=(1,2\cdots m)$.
$$I =\int_{0}^\infty \int_{0}^\infty \left(x+y\right) ^{k} e^{-c(x+y)}d{x}d{y}$$
So I solved using a variable substitution but not polar coordinates.
let $u = x+y$. let $v=x-y$
Therefore $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$
$\frac{\partial{x}}{\partial{u}} = \frac{1}{2}$
$\frac{\partial{x}}{\partial{v}} = \frac{1}{2}$
$\frac{\partial{y}}{\partial{u}} = \frac{1}{2}$
$\frac{\partial{y}}{\partial{v}} = -\frac{1}{2}$
The jacobian is $\frac{\partial{x}}{\partial{u}}\frac{\partial{y}}{\partial{v}} - \frac{\partial{x}}{\partial{v}}\frac{\partial{y}}{\partial{u}} = -1/4 - 1/4 = -\frac{1}{2}$
$$I =\int_{0}^\infty \int_{-u}^u u^{k}e^{-cu}(\frac{1}{2})dvdu$$
$$I =\int_{0}^\infty u^{k+1}e^{-cu}du$$
For this next part I had to use the solution here:
Show that $\int_{0}^{\infty}x^ne^{-x}\,dx=n!$ by differentiating the equality $\int_{0}^{\infty}e^{-tx}\,dx=\frac{1}{t}$
to get a final answer of
$$\frac{(k+1)!}{c^{k+2}}$$
EDIT:
I'm stuck on the generalization.