Integral on part of n-sphere Let $S^n$ be the $n$-sphere and $0&lt;c&lt;1$. Show that $$ \int_{\{x \in S^n |  c\le x^2_1+x^2_2\}} \ln \left (\frac{1}{\sqrt{1-x_1^2 -x^2_2}}-1\right ) dx &lt; \infty$$

Since we are integrating over the n-sphere I'm not sure how to calculate this.
some idea:It is well konw n-sphere
$$\int_{x^2_{1}+x^2_{2}+\cdots+x^2_{n}\le a^2} dx_{1}dx_{2}\cdots dx_{n}=\dfrac{2\pi^{n/2} a^n}{n\Gamma({n/2})},a&gt;0$$
so
$$I=\int_{x\in S^n|c\le x^2_{1}+x^2_{2}}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}dx=\int_{x^2_{1}+x^2_{2}\ge c}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}\dfrac{2\pi^{(n-2)/2}[1-(x^2_{1}+x^2_{2})]^{(n-2)/2}}{(n-2)\Gamma{((n-2)/2})}dx_{1}dx_{2}$$
so you only can prove
$$J=\int_{x^2_{1}+x^2_{2}\ge c}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}\cdot [1-(x^2_{1}+x^2_{2})]^{(n-2)/2}dx_{1}dx_{2}&lt;+\infty$$
let $$x_{1}=r\cos{t},x_{2}=r\sin{t},\sqrt{c}\le r&lt;1$$
then
$$J=2\pi\int_{\sqrt{c}}^{1}r[1-r^2]^{(n-2)/2}\ln{\left(\dfrac{1}{\sqrt{1-r^2}}-1\right)}&lt;+\infty$$
$$\Longleftrightarrow 
K=\int_{c}^{1}(1-u)^{(n-2)/2}\ln{\left(\dfrac{1}{\sqrt{1-u}}-1\right)}du&lt;+\infty,0&lt;c&lt;1$$
let
$\sqrt{1-u}=t$,then
$$\Longleftrightarrow \int_{0}^{\sqrt{1-c}}t^{n-1}\ln{\left(\dfrac{1}{t}-1\right)}dt&lt;\infty$$
then I think you can do it?
some idea:It is well konw n-sphere
$$\int_{x^2_{1}+x^2_{2}+\cdots+x^2_{n}\le a^2} dx_{1}dx_{2}\cdots dx_{n}=\dfrac{2\pi^{n/2} a^n}{n\Gamma({n/2})},a&gt;0$$
so
$$I=\int_{x\in S^n|c\le x^2_{1}+x^2_{2}}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}dx=\int_{x^2_{1}+x^2_{2}\ge c}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}\dfrac{2\pi^{(n-2)/2}[1-(x^2_{1}+x^2_{2})]^{(n-2)/2}}{(n-2)\Gamma{((n-2)/2})}dx_{1}dx_{2}$$
so you only can prove
$$J=\int_{x^2_{1}+x^2_{2}\ge c}\ln{\left(\dfrac{1}{\sqrt{1-x^2_{1}-x^2_{2}}}-1\right)}\cdot [1-(x^2_{1}+x^2_{2})]^{(n-2)/2}dx_{1}dx_{2}&lt;+\infty$$
let $$x_{1}=r\cos{t},x_{2}=r\sin{t},\sqrt{c}\le r&lt;1$$
then
$$J=2\pi\int_{\sqrt{c}}^{1}r[1-r^2]^{(n-2)/2}\ln{\left(\dfrac{1}{\sqrt{1-r^2}}-1\right)}&lt;+\infty$$
$$\Longleftrightarrow 
K=\int_{c}^{1}(1-u)^{(n-2)/2}\ln{\left(\dfrac{1}{\sqrt{1-u}}-1\right)}du&lt;+\infty,0&lt;c&lt;1$$
let
$\sqrt{1-u}=t$,then
$$\Longleftrightarrow \int_{0}^{\sqrt{1-c}}t^{n-1}\ln{\left(\dfrac{1}{t}-1\right)}dt&lt;\infty$$
then I think you can do it?