# Integral $\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi$

Hi I am trying to prove this result below $$\mathcal{J}:=\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi.$$ The result is quite interesting however I realized I am not familiar with working square roots of log functions like this. I am more stumped than usual because there isn't much to work with here. The indefinite integral is given by $$\int \frac{dx}{\sqrt{\log \frac{1}{x}}}=-\sqrt \pi\, \text{erf}\left(\sqrt {\log \frac{1}{x}}\right),$$ although I cannot prove this. Possibly a proof to this will lead to the result of the definite integral.

It seems the integral is somehow related to a Gaussian integral possibly, I notice the error function and the result $\sqrt \pi$. A solution would be greatly appreciated and I hope could also be of use to the math community here.

Thanks. In case anybody likes this integral and is interested in a similar one, here is another one for you:

$$\int_0^1 \sqrt{\log \frac{1}{x}} \,dx=\frac{\sqrt \pi}{2}$$

• I think these are Gamma function values
– MPW
May 26 '14 at 4:19
• @MPW yes exactly it sure seems like that May 26 '14 at 4:22
• Ah ok, but I have got a solution to that, I will post it anyways. :) May 26 '14 at 4:24
• Taking $u=\sqrt{\ln\dfrac{1}{x}}$ would solve the integral. May 26 '14 at 4:25
• @Tunk-Fey Thank you my friend :) May 26 '14 at 4:26

Rewrite $\log(1/x)=-\log(x)$ and use the substitution $-\log x=t^2 \Rightarrow dx=-2te^{-t^2}dt$ to obtain: $$I=\int_0^1 \frac{dx}{\sqrt{\log \frac{1}{x}}}=\int_0^{\infty} 2e^{-t^2}\,dt=\boxed{\sqrt{\pi}}$$

For the second one, do the same thing to obtain the integral: $$\int_0^{\infty} 2t^2e^{-t^2}\,dt$$ Since $$\int_0^{\infty} e^{-(at)^2}\,dt=\frac{\sqrt{\pi}}{2a}$$ Differentiate both the sides wrt $a$ to obtain: $$\int_0^{\infty} 2at^2e^{-(at)^2}\,dt=\frac{\sqrt{\pi}}{2a^2}$$ Substitute $a=1$ to obtain the answer.

• I am going to accept the answer in a minute when the system lets me! :) +1 Thank you my friend May 26 '14 at 4:26
• Thanks but this integral is far more easier than the other ones you have posted here. :P May 26 '14 at 4:29
• I do not always try to post very difficult integrals, I post integrals that I think people may find interesting. Difficult or not, they are all beautify :)Thanks I just checked this as the answer, sorry I forgot to yesterday May 26 '14 at 21:44

This is Euler's first historical integral expression for the $\Gamma$ function. No substitution necessary. The integral is simply $\Big(-\frac12\Big)!=\Gamma\Big(1-\frac12\Big)=\Gamma\Big(\frac12\Big)=\sqrt\pi.~$ QED.

• Yes, that's how I recognized it, thanks for the info. But I think most people today find the form I mentioned most familiar.
– MPW
May 26 '14 at 12:51
• excellent to know, I never knew this! Thanks a lot @Lucian May 26 '14 at 21:46

Note that $$\int_0^1\left(\log\frac1x \right)^pdx=\int_0^{\infty}u^pe^{-u}du=\Gamma(p+1)$$ by making the simple substitution $x=e^{-u}$.

$$\int_0^1 \sqrt{\log \frac{1}{x}} \,dx= x \sqrt{\log \frac{1}{x}}\big{|}_0^1- \int_0^1 x \frac{dx}{(-2x)\sqrt{\log \frac{1}{x}}}\\ =\frac12 \frac{}{}\int_0^1 \frac{dx}{\sqrt{\log \frac{1}{x}}}\\ =\frac{\sqrt{\pi}}{2}$$