MHB Ready to Tackle an Advanced Calculus Challenge?

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The discussion revolves around evaluating the integral $$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta$$ and demonstrating its equivalence to another integral form. Participants highlight the use of transformations and substitutions, particularly involving the tangent function and elliptic integrals, to simplify the problem. The solution ultimately connects to known results involving the Gamma function and elliptic integrals, revealing a complex relationship between these mathematical entities. A modified version of the integral is proposed for further exploration, adding to the challenge. The conversation showcases advanced techniques in calculus and integral evaluation.
Shobhit
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$$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \frac{\log 2}{16 \Gamma \left(\frac{3}{4} \right)^2}\sqrt{2\pi^3}$$

This integral is harder than the http://mathhelpboards.com/challenge-questions-puzzles-28/integration-challenge-7720.html. :D
 
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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
 
Not surprised to see elliptic integrals and hypergeometric functions involved. I tried to solve it with no success.
 
Well done RV! :)

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

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