Lebesgue integral exist

Question

They ask me to determine the $ p $ values ​​for which the following Lebesgue integrals exist $$ \int_0 ^ {\infty} x ^ x e ^ {- x^{p} } dx$$ I have to see when it is finite, but the analysis is quite complicated mainly because in all my degree I do not deal with the integral of $x ^ x$ so I investigated but it is not clear to me until the moment of doing some calculations I have to $ p $ in $ [1, \infty) $ but in the other cases I don't know how to proceed, if you have any advice or help I would appreciate it.

Answer 1

The original question is $\displaystyle\int_{0}^{\infty}x^{x}e^{-xp}dx$:

For any $p$, $\lim_{x\rightarrow\infty}x^{x}e^{-xp}dx=\infty$, there is no such $p$ for which the integral exists as a real number.

For the new question, for $p>1$, first we observe that \begin{align*} \log\dfrac{x^{x}}{e^{x^{p}}}=x\log x-x^{p}\leq C_{q}x^{q}-x^{p}, \end{align*} for some $1<q<p$ and constant $C_{q}>0$, so \begin{align*} \dfrac{x^{x}}{e^{x^{p}}}\leq e^{C_{q}x^{q}-x^{p}}. \end{align*} But now we compare with $e^{-x^{p}/2}$: \begin{align*} \dfrac{e^{C_{q}x^{q}-x^{p}}}{e^{-x^{p}/2}}\rightarrow 0. \end{align*} To see this comparison, note that \begin{align*} \log\dfrac{e^{C_{q}x^{q}-x^{p}}}{e^{-x^{p}/2}}=C_{q}x^{q}-x^{p}+\dfrac{x^{p}}{2}=x^{q}\left(C_{q}-\dfrac{1}{2}x^{p-q}\right)\rightarrow-\infty \end{align*} as $x\rightarrow\infty$.

But it is now easy to see that \begin{align*} \int_{0}^{\infty}e^{-x^{p}/2}dx<\infty, \end{align*} so the integral \begin{align*} \int_{0}^{\infty}x^{x}e^{-x^{p}}dx<\infty. \end{align*} Keep in mind that we start with $p>1$.

But for $p\leq 1$, we have $x^{x}e^{-x^{p}}\rightarrow\infty$. To see this, once again we take the $\log$ to have \begin{align*} x\log x-x^{p}=x^{p}\left(x^{1-p}\log x-1\right)\rightarrow\infty. \end{align*} So the integral in this case cannot exist as a real number.