I am practicing to integrate with the incomplete gamma function. As such, I want to integrate $$\int\ln^e(x)dx$$ I use the following substitutions $$u=\ln(x)\implies \int u^e e^u du$$ $$v=-u\implies \int(-v)^e e^{-v}\cdot -dv$$ Now, we can convert this to a gamma function $$\int(-v)^e e^{-v}\cdot -dv = - \int(-1)^e v^e e^{-v} dv= -(-1)^e\int v^{(1+e)-1} e^{-v} dv = -(-1)^e\cdot -\Gamma(1+e, -\ln(x)) $$ which simplifies to $$(-1)^e\cdot \Gamma(1+e, -\ln(x)) +C$$ This is wrong though, and wolfram alpha gives me an answer of $${\color{red}{(-1)^{-e}}}\cdot \Gamma(1+e, -\ln(x)) +C$$
Something clearly went wrong here (and here, for that matter since I somehow also have a missing negative from the power?) and I'm not exactly sure what went wrong.
I conjecture that perhaps splitting $(-v)^e = (-1)^e v^e$ was incorrect since the interior is negative, but I am not sure what to do otherwise with such a term.
Can anyone enlighten me as to what my mistake is and how I would properly integrate this function?
Thanks
Edit: Okay, differentiating the results seem to provide interesting results. Differentiating my own result on wolfram alpha gives $$(-1)^e (-\ln(x))^e$$ which equals the initial function if i combine the powers, while the negative power does not work. I find this odd.
I again conjecture that since the inner parts of the power functions are negative, I violated this rule when integrating and violating the rule again here reverts it. I'm still not sure what to do with the original integral though.
Let $0<y<x$. Then you find
$$\begin{aligned} \int_y^x (\ln(s))^e ds&=\int_{\ln(y)}^{\ln(x)}s^ee^sds\\ &=(-1)^{-e}\int_{\ln(y)}^{\ln(x)}(-s)^ee^sds\\ &=(-1)^{-e}(-1)\int_{\ln(y)}^{\ln(x)}(-s)^ee^s(-1)ds\\ &=(-1)^{-e}(-1)\int_{-\ln(y)}^{-\ln(x)}s^ee^{-s}ds\\ &=(-1)^{-e}\int_{-\ln(x)}^{-\ln(y)}s^ee^{-s}ds. \end{aligned}$$
You find $-\log(y)\rightarrow \infty$ for $y\rightarrow 0$, such that
$$\int_0^x(\ln(s))^e ds=(-1)^{-e}\int_{-\log(x)}^{\infty}s^ee^{-s}ds=(-1)^{-e}\Gamma(1+e,-\ln(x)).$$