MHB A few tricky integrals.... Or are they?

[DreamWeaver]

Here are a few Vardi-type integrals I recently posted on another forum (some of you might have seen them)...Assuming the following classic result - due to Vardi - holds...\int_{\pi/4}^{\pi/2}\log\log(\tan x)\,dx=\frac{\pi}{2}\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]Prove that:
\int_{\pi/4}^{\pi/2}\log^2[\log(\tan x)]\,dx=

\beta''(1)+\frac{\pi^3}{24}-\frac{\pi\gamma^2}{4}-\pi\gamma\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]and...\int_{\pi/4}^{\pi/2}\log^3[\log(\tan x)]\,dx=

\beta'''(1)-3\gamma\,\beta''(1)+\frac{\pi\gamma^3}{2}-\frac{\pi}{2}\zeta(3) + \left(\frac{\pi^3}{4}+\frac{3\pi\gamma^2}{2}\right)\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]
Where $$\beta(x)\,$$ is the Dirichlet Beta function, defined by:$$\beta(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}$$

 

[polygamma]

I saw your other post, so I'm not going to say much.

But here's a brief outline of a way to evaluate Vardi's integral.Let $ \displaystyle I(a) = \int_{0}^{\infty}\frac{\ln(a^{2}+x^{2})}{\cosh x} \ dx $ and differentiate inside of the integral to get $ \displaystyle I'(a) = \int_{0}^{\infty} \frac{2a}{(a^{2}+x^{2}) \cosh x} \ dx $.

$I'(a)$ can be evaluated be letting $ \displaystyle f(z) = \frac{2a}{(a^{2}+z^{2}) \cosh z}$ and integrating around a rectangle or circle in the upper-half complex plane.

After somewhat tedious calculations, you''ll find that $ \displaystyle I'(a) = \Big[ \psi \Big(\frac{3}{4} + \frac{a}{2 \pi} \Big) - \psi \Big(\frac{1}{4} + \frac{a}{2 \pi} \Big) \Big]$

Then integrate back with respect to $a$.

Finding the constant of integration is tricky. It involves rewriting the integral and letting $a$ go to $\infty$.

But it will turn out that $ \displaystyle I(a) = 2 \pi \ln \Bigg[ \sqrt{2 \pi} \frac{\Gamma(\frac{3}{4} + \frac{a}{2 \pi})}{\Gamma(\frac{1}{4} + \frac{a}{2 \pi})} \Bigg] $

Then $ \displaystyle \lim_{a \to 0^{+}} I(a) = 2 \int_{0}^{\infty} \frac{\ln x}{\cosh x} \ dx = 2 \pi \ln \Bigg[ \sqrt{2 \pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} \Bigg] $

Finally make the change of variables $x = \ln (\tan u) $.

 

[DreamWeaver]

Very nice, RV! (Rock)

 

[DreamWeaver]

No worries if not, but just in case anyone's interested, here's a little hint to get you started in the right direction...

(Bandit)

Spoiler

Consider the integral:$$\int_0^{\infty}\frac{x^{q-1}}{\cosh x}\,dx$$for the real parameter $$q\in\mathbb{R}^+$$