Welcome to our community

Be a part of something great, join today!

A few tricky integrals... Or are they?

DreamWeaver

Well-known member
Sep 16, 2013
337
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...


[tex]\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][/tex]


Prove that:



[tex]\int_{\pi/4}^{\pi/2}\log^2[\log(\tan x)]\,dx=[/tex]

[tex]\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][/tex]


and...


[tex]\int_{\pi/4}^{\pi/2}\log^3[\log(\tan x)]\,dx=[/tex]

[tex]\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][/tex]



Where \(\displaystyle \beta(x)\,\) is the Dirichlet Beta function, defined by:


\(\displaystyle \beta(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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) $.
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
Very nice, RV! (Rock)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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)


Consider the integral:


\(\displaystyle \int_0^{\infty}\frac{x^{q-1}}{\cosh x}\,dx\)


for the real parameter \(\displaystyle q\in\mathbb{R}^+\)