# How to prove $\int_{B(0,1)}\frac {1}{log\left( 1+\frac{1}{|x|}\right) }dx=C\int_{0}^{1}\frac {1}{\left[log(1+\frac{1}{r})\right]^n}\frac{1}{r}dr$

I was reading a textbook, and in the proof of one of the theorems; the author claims the following without providing the intermediate steps that lead him to make such a claim: $$\int_{B(0,1)}\frac {1}{log\left( 1+\frac{1}{|x|}\right) }dx\leq C\int_{0}^{1}\frac {1}{\left[log(1+\frac{1}{r})\right]^n}\frac{1}{r}dr$$

(I am not sure if it should be $\leq$ or strict equality $=$). Where $B(0,1)$ is the $n-th$ dimensional open unit ball. $x=(x_1,...,x_n)$. $C$ is just a constant.

I tried to switch to polar coordinates in n-dimensional space, but I can't seem to have the expression the author had. Any help is much appreciated!

## 1 Answer

Using the $n$-sphere transformation to transform the integral to $$I=C\int_{0}^1 \frac{r^{n-1}}{\log\left(1+\frac{1}{r}\right)}dr$$ where $$C=\idotsint\limits_V \sin^{N-2}\theta_1\sin^{N-2}\theta_2\cdots\sin\theta_{N-2}d\theta_1d\theta_2\cdots d\theta_{N-2}$$ where $$V=\{0\le \theta_1\le 2\pi,\ 0\le \theta_i\le \pi,\ i=2,\cdots,n-2\}$$