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

#### DreamWeaver

##### Well-known member
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 $$\displaystyle \beta(x)\,$$ 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
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 $x = \ln (\tan u)$.

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Very nice, RV!

#### DreamWeaver

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

Consider the integral:

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

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