I am stuck on showing fornally that the following integral is divergent:
for $n\in \mathbb{N}^+$, $\displaystyle\int_0^1 \frac{r^{n-1}}{(\lambda^2 + r^2)^{n/2}} dr $.
Intuitively, if I take out $\lambda^2$ in the denominator, I have $\displaystyle\int_0^1 \frac{r^{n-1}}{{(r^2)}^{n/2}} dr =\int_0^1 \frac{1}{r} dr$ which clearly diverges...
But how do I reconcile this with the above problem? The tricky part is that $\frac{1}{(\lambda^2 + r^2)^{n/2}} < \frac{1}{r^n}$ and so it is the other direction.
EDIT:
Alright, what if $\lambda \rightarrow 0$? if I take $\lim_{\lambda \rightarrow 0}\displaystyle\int_0^1 \frac{r^{n-1}}{(\lambda^2 + r^2)^{n/2}} dr $?
I know that I can only pass a limit inside an integral if the integrand is continuous in a closed domain. So if I am considering $(\lambda,r) \in [0,1]\times[0,1]$, the integrand is not continuous at $(0,0)$. any ideas how to reconcile this?
If $\lambda\ne 0$, the integral converges, since our function is continuous in the closed interval $[0,1]$.
Edit: For the modified problem, assume $\lambda$ is positive, and make the change of variable $r=\lambda t$. We end up with $$\int_0^{1/\lambda} \frac{t^{n-1}}{(1+t^2)^{n/2}}\,dt.$$
For large $t$, the integrand behaves like $1/t$, so the integral blows up as $\lambda\to 0^+$.