I am interested in alternative solutions to the following question:
Prove that $\int_{0}^{\infty}x^{-\ln x}dx=\sqrt{\pi\sqrt{e}}$
I first used $\int_{0}^{\infty}x^{-\ln x}dx=\int_{0}^{\infty}e^{-\ln^{2}x}dx$
Then I used a u-sub to get:
$$\frac{1}{2}\int_{0}^{1}\left(e^{\sqrt{-\ln u}}+e^{-\sqrt{-\ln u}}\right)\frac{ue^{\ln^{2}\left(e^{-\sqrt{-\ln u}}\right)}}{\sqrt{-\ln u}}du$$
I realized I could simplify it to $\frac{1}{2}\int_{0}^{1}\frac{e^{\sqrt{-\ln u}}+e^{-\sqrt{-\ln u}}}{\sqrt{-\ln u}}du$.
After another substitution, I got $\int_{0}^{\infty}\left(e^{v}+e^{-v}\right)\left(e^{-v^{2}}\right)dv$
I solved using $\text{erfc}(x)$:
$$\int_{0}^{\infty}\frac{e^{2v}+1}{e^{v+v^{2}}}dv=\int_{0}^{\infty}\left(e^{v-v^{2}}+e^{-v-v^{2}}\right)dv=\sqrt{\pi\sqrt{e}}$$
I feel like there should be a better way to do this.
What other methods can be used to solve this?
Best Answer
Well, substitute $\ln(x) = y$, then $\int_{0}^{\infty} x^{-\ln(x)} dx = \int_{-\infty}^{\infty} e^{y}e^{-y^2}dy = e^{1/4}\int_{-\infty}^{\infty} e^{-t^2} dt = e^{1/4}\sqrt{\pi}$.