Please help me to handle the following problem:
Prove that if $\alpha>0$, then $$f(\alpha)=\int_0^{+\infty} \frac{x}{x^{\alpha+2}+1}dx$$
is differentiable.
My attempt:
I am going to use the following fact:
If $f(x,\alpha)$ and $f'_{\alpha}(x,\alpha)$ are continuous in $[0,+\infty)\times(0, +\infty)]$ and
$I(\alpha)=\int_0^{+\infty}f(x,\alpha)$ is convergent
$\int_0^{+\infty}f'_{\alpha}(x,\alpha)$ is convergent uniformly for $\alpha \in (0,+\infty)$
then for any $\alpha \in (0,+\infty)$ there exists $I'(\alpha)=\int_0^{+\infty}f'_{\alpha}(x,\alpha)$
In my notes this is called Leibniz rule, but I do not know standard name of this result in English. Please let me know.
In fact this result was mentioned in a bit more general settings, but I put constants (integration limits) related to the problem.
We have $f'_{\alpha}(x,\alpha)=-\frac{x^{\alpha+3}\log x}{(x^{\alpha+2}+1)^2}$. So $f(x,\alpha)$ is continuous $[0,+\infty)\times(0, +\infty)]$, but $f'_{\alpha}(x,\alpha)$ is not, because it is not defined when $x=0$. How should I handle this issue?
$$f(\alpha)=\int_0^{+\infty} \frac{x}{x^{\alpha+2}+1}dx < \int_0^{+\infty} \frac{x}{x^{\alpha+2}}dx=\int_0^{+\infty} \frac{1}{x^{\alpha+1}}dx<+\infty$$ This is well know fact.
Now I have to prove that that $$f'_{\alpha}(x,\alpha)=-\frac{x^{\alpha+3}\log x}{(x^{\alpha+2}+1)^2}$$ converges uniformly. If there were no $\log x$ factor, that would be easy, but it is not the case. How should I prove uniform convergence?
Thanks a lot for your answers and for pointing on my mistakes.