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 \(\displaystyle \beta(x)\,\) is the Dirichlet Beta function, defined by:\(\displaystyle \beta(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}\)

 

[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:\(\displaystyle \int_0^{\infty}\frac{x^{q-1}}{\cosh x}\,dx\)for the real parameter \(\displaystyle q\in\mathbb{R}^+\)

 

[CellCoree]

I find these integrals to be quite interesting and challenging. The result by Vardi, which is used as a starting point, is a well-known and important result in mathematics. It relates the logarithmic integral function to the gamma function, which has many applications in various fields of science.

To prove the first integral, we can use the substitution u = log(tan(x)). This gives us du = sec^2(x) dx, and the limits of integration become log(sqrt(2)) and log(1). Using the result by Vardi, we can write the integral as:

int_{log(sqrt(2))}^{log(1)} u^2 du = \beta''(1) = \frac{\pi}{2}\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]

Hence, we have proved the first integral by using the known result for \beta''(1).

Similarly, for the second integral, we can use the substitution u = log(tan(x)) again. This gives us du = sec^2(x) dx and the limits of integration become log(sqrt(2)) and log(1). Using the result for \beta'''(1) and the result by Vardi, we can write the integral as:

int_{log(sqrt(2))}^{log(1)} u^3 du = \beta'''(1) = \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]

This proves the second integral as well.

The third integral is more complex and requires the use of multiple properties of the Dirichlet Beta function and the gamma function. We can use the same substitution as before and expand u^3 in terms of \beta(x) to get:

int_{log(sqrt(2))}^{log(1)} u^3 du = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}\int_{log(sqrt(2))}^{log(1)}sec^2(x)dx

Using the known result for \beta''(1) and the integral