Ready to Tackle an Advanced Calculus Challenge?

[Shobhit]

Show that

$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \frac{\log 2}{16 \Gamma \left(\frac{3}{4} \right)^2}\sqrt{2\pi^3}$$

This integral is harder than the http://mathhelpboards.com/challenge-questions-puzzles-28/integration-challenge-7720.html. :D

 

[polygamma]

I'm just going to show that it is equivalent to another definite integral.
$$ \int_0^{\pi /2}\frac{\log ( \tan x) }{\sqrt{1+\cos^2 x }}\ dx = \int_{0}^{\pi /2} \frac{\log (\tan x)}{\sqrt{2-\sin^{2} x}} \ dx = \frac{1}{\sqrt{2}} \int_{0}^{\pi /2} \frac{\log(\tan x)}{\sqrt{1- \frac{1}{2} \sin^{2} x}} \ dx $$Let $ u = \tan x$.$$ = \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{\log u}{\sqrt{1- \frac{1}{2} \frac{u^{2}}{1+u^{2}}}} \frac{1}{1+u^{2}} \ du = \int_{0}^{\infty} \frac{\log u}{\sqrt{\frac{2+u^{2}}{1+u^{2}}}} \frac{1}{1+u^{2}} \ du = $$

$$ = \int_{0}^{\infty} \frac{\log u}{\sqrt{(1+u^{2})(2+u^{2})}} \ du = \int_{0}^{\infty} \frac{\log u}{\sqrt{(1+u^{2})(1 - i^2+u^{2})}} \ du $$There is a formula that states $$ \int_{0}^{\infty} \frac{\log x}{\sqrt{(1+x^{2})(1-k^{2} + x^{2})}} \ dx = \frac{1}{2} K(k) \ln( \sqrt{1-k^{2}})$$

where $K(k)$ is the complete elliptic integral of the first kind.A derivation in one of Victor Moll's papers uses a crazy-looking hypergeometric identity.So anyways

$$ \int_0^{\pi /2}\frac{\log ( \tan x) }{\sqrt{1+\cos^2 x }}\ dx = \frac{1}{2} K(i) \ln (\sqrt{2}) = \frac{\log 2}{16 \sqrt{2 \pi}} \Gamma^{2} \left( \frac{1}{4} \right)$$

which by the Gamma reflection formula is equivalent to the answer given

 

[alyafey22]

Not surprised to see elliptic integrals and hypergeometric functions involved. I tried to solve it with no success.

 

[Shobhit]

Well done RV! :)

Now, I am going to modify this problem slightly to make it even more challenging.Show that

$$
\int_0^{\pi\over 2}\frac{\log(\tan x)}{\sqrt{2} \sin(x)+\sqrt{1+\sin^2 x}}dx = \frac{1}{\sqrt{2\,\pi}}\left(1+\frac{\log 2}{4} \right)\Gamma\left(\frac34\right)^2-\frac{\sqrt{2\,\pi^3}}{8\Gamma\left(\frac34\right)^{2}}+(\log 2-1)\,\sqrt2
$$

 

[CellCoree]

I am always up for a challenge and I love solving complex problems. Let's take a closer look at this integral and see if we can find a solution.

First, let's rewrite the integral using the properties of logarithms and trigonometric functions:

$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \int_0^{\frac{\pi}{2}}\frac{\log \sin \theta - \log \cos \theta}{\sqrt{1+\cos^2 \theta}}d\theta$$

Next, we can use the substitution $u = \sin \theta$, which gives us $du = \cos \theta d\theta$ and changes the limits of integration to $0$ and $1$:

$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \int_0^1\frac{\log u}{\sqrt{1+1-u^2}}du - \int_0^1\frac{\log (1-u)}{\sqrt{1+1-u^2}}du$$

We can now use the Beta function to evaluate these integrals:

$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \frac{1}{2}\cdot B\left(\frac{1}{2}, \frac{1}{4}\right)-\frac{1}{2}\cdot B\left(\frac{1}{2}, \frac{3}{4}\right)$$

Using the properties of the Beta function, we can simplify this to:

$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \frac{1}{2}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}-\frac{1}{2}\cdot \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left