Compute $\int_0^\infty \left(\frac{1}{x}\right)^{\ln(x)}\,\mathrm{d}x$ [closed]

Solution 1:

It follows from the following reasoning:

Let $u = \log x \implies dx = e^u du$

$$\int_0^\infty x^{-\log x}\:dx = \int_{-\infty}^\infty (e^u)^{-u} e^u \:du = \int_{-\infty}^\infty e^{-u^2 + u}du$$

$$ = \int_{-\infty}^\infty e^{-\left(u-\frac{1}{2}\right)^2+\frac{1}{4}}du = e^{\frac{1}{4}} \int_{-\infty}^\infty e^{-v^2}dv = e^{\frac{1}{4}} \sqrt{\pi}$$

by translating and then using the known value of the Gaussian integral.

Solution 2:

Substitute $x=e^{-t}$. That gives you $dx=-e^{-t}dt$. Hence, integral becomes $\int\limits_{\infty}^{-\infty}-(e^t)^{-t}e^{-t}dt$. Changin the integral to $\int\limits_{-\infty}^{\infty}e^{-t^2-t-0.25}e^{0.25}dt$. Substitute $t+0.5=v$ and the rest is standard.