Domain of definition of $\Gamma_1(x) = \int_0^{+\infty} e^{-t} t^x \ln t dt$

Question

Let

$\displaystyle \Gamma_0(x) = \int_0^{+\infty} e^{-t} t^x dt$

$\displaystyle \Gamma_1(x) = \int_0^{+\infty} e^{-t} t^x \ln t dt$

$\displaystyle \Gamma_2(x) = \int_0^{+\infty} e^{-t} t^x (\ln t)^2 dt$

I want to find the domain of definition of these three functions.

I have found $]-1,+\infty[$ for the first one.

There is no problem around $+\infty$ but there is around $0^+$ for the three functions.

For $x>-1, e^{-t}t^x \ln t \sim_0 t^x \ln t = o(\frac{1}{t^\alpha})$ where $\alpha \in ]0,1[$ is chosen such that $\alpha +x>0$, so $\Gamma_1$ is defined on $]-1,+\infty[$ with Riemann, but I don't know how to prove that it is (or not) defined on $]-\infty,-1]$.

I assume it will be the same trick for $\Gamma_2$?

Answer 1

In general, we have

$$-\ln(x)<\frac1{ax^a}$$

which holds for all $a>0$ for $x>0$. A quick sketch of the proof of such can easily be done by noting that

$\lim_{x\to0}\frac{\ln(x)}{1/ax^a}=0$

$f(x)=\frac1{ax^a}+\ln(x)$ has one point such that $f'(x)=0$ within it's domain, at $x=1$, and is monotone on each side of it.

Now, we have

$$\left|\int_0^1e^{-t}t^x(\ln(t))^n\ dt\right|<\int_0^1e^{-t}t^x\frac1{a^nt^{an}}\ dt=\frac1{a^n}\int_0^1e^{-t}t^{x-an}\ dt$$

Since we can choose $a$ to be close enough to $0$ such that $x-an>-1$, it converges for all $x>-1$.

Trivially, it does not converge at $x\le-1$ since you can compare it to $\Gamma_0$.