# The Log-Trig Integral

The following is the reposted version of Kartik Sharma's problem on logarithmo-trigonometric integral:

If${ R }_{ n }^{ + }=\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt [ { 2 }^{ n } ]{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +..+\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { { \left( \ln { \cos { x } } \right) }^{ 2 } }{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } } } } } } dx$then evaluate $${ R }_{ n }^{ + }$$

Here is the solution to the problem:

In this case..first we observe that the nested radicals are of the form:$\sqrt { \frac { 1 }{ 2 } +\frac { a }{ 2 } }$ which suggests that $a$ can be considered as $\cos { \theta }$ for some $\theta$..Thus the nested radical part of the integral reduces to $\sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +...+\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { \cos { \theta } }{ 2 } } } } } =\cos { \left( \frac { \theta }{ { 2 }^{ n } } \right) }$ where $\cos { \theta } =\sqrt { \frac { { \left( \ln { \cos { x } } \right) }^{ 2 } }{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } } \\ \Rightarrow \sec ^{ 2 }{ \theta } =\frac { { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } }{ { \left( \ln { \cos { x } } \right) }^{ 2 } } =1+\frac { { x }^{ 2 } }{ { \left( \ln { \cos { x } } \right) }^{ 2 } } =1+\tan ^{ 2 }{ \theta } \\ \Rightarrow \tan { \theta } =\pm \frac { x }{ \ln { \cos { x } } }$ We choose $\theta$ to be such that $\tan { \theta } =- \frac { x }{ \ln { \cos { x } } }$. Now, let $r\in \left( 0,1 \right)$. We use the two following results:$\int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( ay \right) } { e }^{ -by } } dy=\Gamma \left( r \right) \frac { \cos { \left\{ r\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } \right\} } }{ { \left( { a }^{ 2 }+{ b }^{ 2 } \right) }^{ \frac { r }{ 2 } } }$ and $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( yx \right) } \cos ^{ y }{ x } } dx=\frac { \pi }{ { 2 }^{ y+1 } } \$/extract_itex] which are trivial. Then we can write $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { \left\{ r\tan ^{ -1 }{ \left( -\frac { x }{ \ln { \cos { x } } } \right) } \right\} } }{ { \left\{ { x }^{ 2 }+{ \left( \ln { x } \right) }^{ 2 } \right\} }^{ \frac { r }{ 2 } } } } dx=\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left\{ \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dy \right\} } dx\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dy } dx\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dx \right\} } dy\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( xy \right) } \cos ^{ y }{ x } } dx \right\} } dy\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\frac { \pi }{ { 2 }^{ y+1 } } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }{ 2 }^{ -y } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }{ e }^{ -y\ln { 2 } } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } .\frac { \Gamma \left( r \right) }{ { \left( \ln { 2 } \right) }^{ r } } =\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ r } } \\$ (clearly in steps 2 and 3 we have used Fubini's theorem and in step 2 I have used gamma function since $r>0$.) We can apply analytic continuation to the above result to extend the domain of $r$ from $(0,1)$ to $(-1,1)$ since the original domain $(0,1)$ is open in $R$ and the integrand function is analytic in $r$. Now we simply put $r=-\frac { 1 }{ { 2 }^{ n } }$ to get that ${ R }_{ n }^{ + }=\frac { 2 }{ \pi } .\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ -\frac { 1 }{ { 2 }^{ n } } } } ={ \left( \ln { 2 } \right) }^{ { 2 }^{ -n } }$. Note by Kuldeep Guha Mazumder 5 years, 5 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. 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