Given indefinite integral $I = \int_{0}^{\infty} x^{\alpha - x}$.
I need to understand for all possible real values of $\alpha$ its' convergence.
I have hypothesis that it converges for $\alpha > -1$. I just drew graph and could clearly see that.
However, what is formal way to show that? When I tried to, I wanted to use Abelian or Dirichlet criterion but they both don't seem to be applicable here. Probably I don't see an obvious proof using comparison test or something like that. Also, I had some problems working with $1/x^x$ because it doesn't seem to have a "nice" integral (by "nice" I mean integral that can be expressed via elementary functions).
All help will be appreciated.
We can compare with the integrals $\int_0^1 x^u \,\mathrm{d}x$ and $\int_1^\infty x^u \,\mathrm{d}x$, with constant $u$. We use that $x^u$ is monotonically increasing in $u$ (for $x>0$), so $u < v$ implies $x^u < x^v$.
Recall that $\int_0^1 x^u \,\mathrm{d}x$ diverges if $u \leq -1$ and converges otherwise. For $x \in (0,1)$, $\alpha - x \leq -1$ when $\alpha \leq -1$. (Also, for $-1 < \alpha, 0$, for $1 + \alpha \leq x < 1$, but this does not include a neighborhood of $0$, so does not resolve convergence.) So the portion of your integral on $(0,1]$ diverges for $\alpha \leq -1$ and converges for $\alpha > -1$.
Recall that $\int_1^{+\infty} x^u \,\mathrm{d}x$ diverges if $u \geq -1$ and converges otherwise. For $x \in [1,\infty)$, $\alpha - x < -1$ if $\alpha < 0$ and for $x \in (1+\alpha, \infty)$, $\alpha - x < -1$ if $\alpha \geq 0$. So the portion of your integral on $[1,\infty)$ converges for any $\alpha$.
Therefore, your integral converges for $ \alpha > -1$ and diverges for $\alpha \leq -1$.