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Here is my question:

Starting from the relation $$\int_{0}^{+\infty}t^{a-1}e^{-nt}\,dt=n^{-a}\Gamma(a)\qquad a>0$$ and differentiating $m-$times under the integral sign we can get to

$$\int_{0}^{+\infty}t^{a-1}\ln^mte^{-nt}\,dt=n^{-a}\sum_{t=0}^{m}(-1)^{m-r}\binom{m}{r}\Gamma^{(r)}(a)\ln^{m-r}n\quad a>0,\;m=0,1,2,\ldots\quad (1)$$ where $\Gamma^{(r)}(a)$ means the $r-$th derivative of $\Gamma$ function at $a$.

Is there a way to reach to similar relation for $\mathbb N\not\ni m>0$ ? (with the $\ln$ in absolute value in this case) The relation doesn't have to be an "exact" result (I haven't find it in any table of Laplace transforms) but I would expect an asymptotic estimate for $n\to+\infty$ or sth could be found.

First thought was to write the above sum in $(1)$ as \begin{align}n^{-a}\sum_{r=0}^{m}(-1)^r\binom{m}{r}\Gamma^{(m-r)}(a)\ln^rn&=n^{-a}\sum_{r=0}^{m}\binom{-m+r-1}{r}\Gamma^{(m-r)}(a)\ln^rn\notag \\&=n^{-a}\sum_{r=0}^{m}\binom{-r-1}{m-r}\Gamma^{(r)}(a)\ln^{m-r}n\notag \\&=n^{-a}\sum_{r=0}^{+\infty}\binom{-r-1}{m-r}\Gamma^{(r)}(a)\ln^{m-r}n\notag \\&=n^{-a}\sum_{r=0}^{+\infty}\frac{\Gamma(-r)}{\Gamma(-m)\Gamma(m-r+1)}\Gamma^{(r)}(a)\ln^{m-r}n\notag\end{align}

and try to find an argument to pass from the validity of the equality on the positive integers to positive reals but I didn't succeed.

Second thought was to start doing repeated integration by parts on LHS of $(1)$. This way I managed to get the first summand but things get really messy and I cannot proceed to get the derivatives of $\Gamma$. Any thoughts?

Thank you very much in advance,

Anastasios

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