How to show that the integral of $\int_{\Bbb R^n }(1 + |x|)^{-L} \mathrm dx$ exist when $L>n$

Question

How to show that the integral of $\int_{\Bbb R^n }(1 + |x|)^{-L} \mathrm dx$ exists (in the sense of Lebesgue integral) when $L>n$? I have computed the case for $\Bbb R^2$ using polar coordinates and I am guessing we can use spherical coordinates for $\Bbb R^3$. Now the problem though is that polar coordinates in higher dimensions get more and more complicated, but since our function is radial, shouldn't there be a much simpler version of polar coordinates we can use? Now if there is a way we can prove the statement without changing coordinates, I would like to know how.

Answer 1

Using hyperspherical coordinates, one can write this integral as $$\int_{\Bbb R^n}\frac{1}{(1+|x|)^{\lambda}}\mathrm d^n x \\ =\underbrace{\int\limits_0^\pi\cdots \int\limits_0^{2\pi}}_{\theta~\text{integrals}}~\int\limits_0^\infty\frac{1}{(1+r)^\lambda} r^{n-1}\mathrm dr~~\text{(sines,cosines)}\mathrm d\theta_1\cdots\mathrm d\theta_{n-1}$$ This can now be written just as the product of the integrals, $$=\underbrace{\int_0^\pi\cdots\int_0^{2\pi}\text{(sines,cosines)}\mathrm d\theta_1\cdots\mathrm d\theta_{n-1}}_{\text{surf. area of }S^{n-1}}~~~\cdot~~~\int_0^\infty\frac{1}{(1+r)^{\lambda}}r^{n-1}\mathrm dr$$ To show convergence, the precise surface area of the $n-1$ sphere is not important, but in any case it is $$A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$ So in all $$I_n(\lambda)=\int_{\Bbb R^n}\frac{1}{(1+|x|)^{\lambda}}\mathrm d^n x=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_0^\infty \frac{1}{(1+r)^\lambda}r^{n-1}\mathrm dr$$ Which is clearly convergent only when $\lambda >n$, and assuming this convergence has the exact value $$I_n(\lambda)=\frac{2\pi^{n/2}}{\Gamma(n/2)}~\mathrm B(n,\lambda-n)$$ Where $\mathrm B$ is the Beta function.

Answer 2

Let $S_n(R)$ be the surface area of the ball with radius $R$ in $\mathbb{R}^n$. Of course $S_n(R)=R^{n-1} S_n(1)$ and

$$ \int_{\mathbb{R}^n}\frac{d\mu}{(1+|x|)^L} = \int_{0}^{+\infty}\frac{S_n(R)}{(1+R)^L}\,dR = S_n(1) \int_{0}^{+\infty}\frac{R^{n-1}}{(1+R)^L}\,dR$$ is clearly convergent iff $L>n$. By Euler's Beta function we have $$ \int_{\mathbb{R}^n}\frac{d\mu}{(1+|x|)^L} = \frac{2\pi^{n/2}}{\Gamma(n/2)}B(n,L-n)=\frac{2\pi^{n/2}\Gamma(n)\Gamma(L-n)}{\Gamma(n/2)\Gamma(L)}.$$