Integral $\int_{\mathbb{R}^{n-1}} \frac{1}{(y_1^2 + y_2^2 +...y_{n-1}^2 + C^2)^\frac{n}{2}} dy = \frac{n\alpha(n)}{2C}$

Question

$$\int_{\mathbb{R}^{n-1}} \frac{1}{(y_1^2 + y_2^2 +...y_{n-1}^2 + C^2)^\frac{n}{2}} dy = \frac{n\alpha(n)}{2C}$$ where $\alpha(n)$ is the volume of the unit ball in $\mathbb{R}^n$.

Could anyone help me with this integral? The proof about Green's function that I am reading says this is a direct calculation, but I couldn't see it. The original equation is $$\frac{2x_n}{n\alpha(n)} \int_{\partial \mathbb{R}_+^n} \frac{1}{|x-y|^n} dS(y) = 1 \text{ for each } x\in \mathbb{R}^n.$$ First I would like to show this for $x = (0,0,..., C)$, and intuitively I know the first $n-1$ component of $x$ will not effect the calculation because $y = (y_1, ...y_{n-1}, 0)$ will range over the entire ${\partial \mathbb{R}_+^n}$.

Thank you very much.

Answer 1

I'm assuming $n > 2$. To compute this integral, it will be useful to use the following properties of the Beta function.

1. $B(x,y) = \int_0^\infty t^{x - 1}(1 + t)^{-x - y}\, dt$, $x > 0, y > 0$.
2. $B(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x + y)$, $x > 0, y > 0$.

Switching the spherical coordinates, we compute

\begin{align}\int_{\Bbb R^{n-1}} \frac{1}{(y_1^2 + \cdots + y_{n-1}^2 + C^2)^{n/2}}\, dy &= \int_0^\infty \int_{S^{n-2}} \frac{r^{n-2}}{(r^2 + C^2)^{n/2}}\, d\omega\, dr \\ &= |S^{n-2}|\int_0^\infty \frac{r^{n-2}}{(r^2 + C^2)^{n/2}}dr\\ &= |S^{n-2}|\int_0^\infty \frac{(uC)^{n-2}}{(u^2C^2 + C^2)^{n/2}}\, C\, du \quad (u = r/C)\\ &= \frac{|S^{n-2}|}{C}\int_0^\infty \frac{u^{n-2}}{(u^2 + 1)^{n/2}}\, du\\ &= \frac{|S^{n-2}|}{C}\int_0^\infty \frac{v^{(n-2)/2}}{(v + 1)^{n/2}}\, \frac{dv}{2v^{1/2}} \quad (v = u^2)\\ &= \frac{|S^{n-2}|}{2C}\int_0^\infty \frac{v^{(n-3)/2}}{(v + 1)^{n/2}}\, dv\\ &= \frac{|S^{n-2}|}{2C}B\left(\frac{n-1}{2}, \frac{1}{2}\right)\\ &= \frac{|S^{n-2}|}{2C}\frac{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}.\\ \end{align}

Since $|S^{n-2}| = 2\pi^{(n-1)/2}/\Gamma\left(\frac{n-1}{2}\right)$ and $\Gamma\left(\frac{1}{2}\right) = \pi^{1/2}$, we have

$$\frac{|S^{n-2}|}{2C}\frac{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)} = \frac{1}{2C}\frac{2\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)} = \frac{n\alpha(n)}{2C},$$ since $\alpha(n) = 2\pi^{n/2}/[n\Gamma\left(\frac{n}{2}\right)]$.