Proving Zwillinger's Formula #686 - Is it Correct?

[dextercioby]

It's formula #686 of Zwillinger's book "CRC Standard Mathematical Tables and Formulae" 31-st edition, 5-th chapter.

He claims that

$$\int_{0}^{1} \frac{dx}{\sqrt{\ln\left(\ln\frac{1}{x}\right)}} =\sqrt{\pi}$$.

Is it correct...? And if so, how does one find/prove something like that...?

Daniel.

 

[StatusX]

That doesn't look right. As x goes from 0 to 1, ln(1/x) goes from infinity to 0, and ln(ln(1/x)) goes from infinity to -infinity, all monotonically. ln(ln(1/x))=0 at x=1/e and so the integral from 0 to 1/e is real and nonzero, while the integral from 1/e to 1 is imaginary and nonzero, so the answer should be complex.

 

[dextercioby]

I could do that myself as well. It looked rather odd that i couldn't find the integral in the bibliographical resources...

Daniel.

 

[George Jones]

It's formula #686 of Zwillinger's book "CRC Standard Mathematical Tables and Formulae" 31-st edition, 5-th chapter.

He claims that

$$\int_{0}^{1} \frac{dx}{\sqrt{\ln\left(\ln\frac{1}{x}\right)}} =\sqrt{\pi}$$.

Is it correct...? And if so, how does one find/prove something like that...?

Daniel.

It seems that there are too many ln's, i.e.,

$$\int_{0}^{1} \frac{dx}{\sqrt{\ln\frac{1}{x}}} = \Gamma \left( \frac{1}{2} \right) =\sqrt{\pi}$$

 

[dextercioby]

Wow, that can be it. It sounds very reasonable.

Daniel.