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.

  • 1
    $\begingroup$ Substitute $u=x^2$ to obtain a Gaussian integral. $\endgroup$ – David H Feb 23 '15 at 9:52
  • $\begingroup$ @DavidH: Oh, I see, thank you. I also figured out how to do it another way. $\endgroup$ – Marcus Emilsson Feb 23 '15 at 9:58

The function $\Gamma(x)$ is defined as

$$\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \,\mathrm{d}t.$$

This general integral below on the left can be transformed in terms of the gamma function with a substitution like so:

$$\int_0^\infty t^{x-1} e^{-bt} \,\mathrm{d}t = \int_0^\infty \left( \frac{u}{b} \right)^{x-1} \frac{e^{-u}}{b} \,\mathrm{d}u = b^{-x} \Gamma(x).$$

This is in the form of the integral in the question. Plugging in the values yields the desired result, $\sqrt{2\pi}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.