# Integration Challenge 2

#### Shobhit

##### Member
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 previous one. #### Random Variable

##### Well-known member
MHB Math Helper
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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Not surprised to see elliptic integrals and hypergeometric functions involved. I tried to solve it with no success.

#### Shobhit

##### Member
Well done RV! Now, I am going to modify this problem slightly to make it even more challenging.

Show that

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