# Integrating $\int_0^1 \sqrt{\frac{\log(1/t)}{t}} \,\mathrm{d}t = \sqrt{2\pi}$

I'd like to evaluate the integral

$$\int_0^1 \sqrt{\frac{\log(1/t)}{t}} \,\mathrm{d}t.$$

I know that the value is $\sqrt{2\pi}$ but I'm not sure how to get there.

I've tried a substitution of $u = \log(1/t)$, which transforms the integral into

$$\int_0^\infty \sqrt{u e^{-u}} \,\mathrm{d}u.$$

This seems easier to deal with. But where do I go from here? I'm not sure.

• Substitute $u=x^2$ to obtain a Gaussian integral. – David H Feb 23 '15 at 9:52
• @DavidH: Oh, I see, thank you. I also figured out how to do it another way. – Marcus Emilsson Feb 23 '15 at 9:58

The function $\Gamma(x)$ is defined as
$$\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \,\mathrm{d}t.$$
$$\int_0^\infty t^{x-1} e^{-bt} \,\mathrm{d}t = \int_0^\infty \left( \frac{u}{b} \right)^{x-1} \frac{e^{-u}}{b} \,\mathrm{d}u = b^{-x} \Gamma(x).$$
This is in the form of the integral in the question. Plugging in the values yields the desired result, $\sqrt{2\pi}$.