12
$\begingroup$

Prove that:

$\displaystyle\int_{\pi/4}^{\pi/2} \ln (\ln(\tan x))dx =\frac{\pi}{2}\ln \left( \frac{\sqrt{2\pi} \Gamma \left(\dfrac{3}{4} \right)}{\Gamma \left(\dfrac{1}{4} \right)}\right)$

I know that the Vardi's Integral can be evaluated in terms of derivatives of Hurwitz Zeta Function. I would like to see a method which uses differentiation under the integral sign.

$\endgroup$
  • 3
    $\begingroup$ You seem to know very well what you want. Did you try anything in this direction? $\endgroup$ – Fabian Jan 24 '13 at 8:11
  • $\begingroup$ Yes, I have evaluated it using Hurwitz Zeta Function but I want to know if there is an alternate approach. $\endgroup$ – Shobhit Bhatnagar Jan 24 '13 at 8:16
  • $\begingroup$ Let $\tan x=1/t$ and check this out: math.stackexchange.com/questions/121545/… $\endgroup$ – L. F. Jan 27 '13 at 3:28
11
$\begingroup$

I'm going to sketch a way to transform the integral into a sum. The sum looks difficult, but it converges and numerically checks with the stated result.

Begin by substituting $u=\log{\tan{x}}$. Then

$$du = \frac{1}{\tan{x}} \sec^2{x} \, dx = \left ( \frac{1}{\tan{x}} + \tan{x} \right ) dx = (e^{-u} + e^u) \, dx $$

and

$$\begin{align} \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) &= \int_0^{\infty} du \: \frac{\log{u}}{e^u + e^{-u}} \\ &= \int_0^{\infty} du \: \frac{e^{-u} \log{u}}{1+e^{-2 u}} \\ &= \int_0^{\infty} du \: e^{-u} \log{u} \sum_{k=0}^{\infty} (-1)^k e^{-2 k u} \end{align}$$

Reverse the order of sum and integral, which is justified by Fubini's Theorem (both sum and integral are absolutely convergent). Then we may write

$$\begin{align} \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) &= \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} du \: e^{-(2 k+1) u} \log{u} \\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \int_0^{\infty} du \: e^{-u} \log{u} - \sum_{k=0}^{\infty} (-1)^k \frac{\log{(2 k+1)}}{2 k+1} \int_0^{\infty} du \: e^{-u} \\ &= -\frac{\pi}{4} \gamma + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\log{(2 k+1)}}{2 k+1} \\ \end{align} $$

where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand sign is known:

$$ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\log{(2 k+1)}}{2 k+1} = \frac{\pi}{4} \gamma + \frac{\pi}{4} \log{\frac{\Gamma{\left ( \frac{3}{4} \right )}^4}{\pi}} $$

Use the fact that

$$\Gamma{\left ( \frac{3}{4} \right )} \Gamma{\left ( \frac{1}{4} \right )} = \sqrt{2} \pi$$

to deduce that

$$ \int_{\pi/4}^{\pi/2} dx \: \ln (\ln(\tan x)) = \frac{\pi}{2} \log{\left [\sqrt{2 \pi} \frac{\Gamma{\left ( \frac{3}{4} \right )}}{\Gamma{\left ( \frac{1}{4} \right )}}\right ]} $$

$\endgroup$
  • 1
    $\begingroup$ This is just great. Thanks! $\endgroup$ – Shobhit Bhatnagar Jan 27 '13 at 11:59
4
$\begingroup$

As a matter of fact, there is a very nice paper by Iaroslav Blagouchine in which the Vardi's integral, as well as numerous integrals akin to it, are treated in details (more correctly they should be called Malmsten's integrals).

$\endgroup$

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.