I have an integral equation $$\int_{\mathbb{R}^{n}}F(x)dx=1$$ for $F(x):=\frac{C}{(1+|x|^{2})^{\frac{n+1}{2}}}$.
And I want to convert the above integral equality to polar coordinates. Am I right in thinking that it looks something like this: $$\int_{0}^{\infty}\int_{\mathbb{S}^{n-1}}F(r\varphi)d\varphi dr=r^{n-1}$$
Edit: Following Lost in a Maze's hint: $$\begin{aligned}\int_{\mathbb{R}^{n}}F(x)dx & =\int_{0}^{\infty}\int_{\partial B(0,r)}g(r)dSdr \\ &=\int_{0}^{\infty}\int_{\partial B(0,r)}\frac{C}{(1+r^{2})^{\frac{n+1}{2}}}dSdr \\ &=\int_{0}^{\infty}\frac{C}{(1+r^{2})^{\frac{n+1}{2}}}\int_{\partial B(0,r)}dSdr \\ &=2C\frac{\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}\int_{0}^{\infty}\frac{r^{n}}{(1+r^{2})^{\frac{n+1}{2}}}dr \end{aligned}$$
Since I am trying to prove that $C=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}$, I must have that the integral is equal to $\frac{1}{2}$, or I have done something wrong?
Let $g(r) := \frac{C}{(1 + r^2)^{\frac{n+1}{2}}}$ for $r > 0$, so $F(x) = g(|x|)$.
Since $F$ is radial, the change to polar coordinates becomes $$ \int_{\mathbb{R}^n}F(x) \,dx = \int_{0}^{\infty} \int_{\partial B(0,r)}g(r)\,d S\, d r $$ where $\partial B(0,r)$ denotes the boundary of the ball of radius $r$ in $\mathbb{R}^n$. Now $g$ only depends on $r$, so this will allow you to compute the surface integral (it is just the surface area of $B(0,r)$), and then subsequently the integral over $r$.