# Calculus Challenge - 1 (Reposted)

$\large R_n^{+}:=\frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\Large {2^n}]{x^2 + \ln^2\!\cos x} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{ \frac{\ln^{2}\!\cos x} {x^2 + \ln^2\! \cos x}}}}}\,dx$

Evaluate $$R_n$$.

###### Previously some calculus challenges were posted by Kartik Sharma. Some of those challenges were unsolved. So I wanted to repost them. And let us see who will stand in the top position in these series of challenges.

Note by Surya Prakash
2 years, 11 months ago

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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 $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( yx \right) } \cos ^{ y }{ x } } dx=\frac { \pi }{ { 2 }^{ y+1 } } \\$ 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 $${ R }_{ n }^{ + }=\frac { 2 }{ \pi } .\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ -\frac { 1 }{ { 2 }^{ n } } } } ={ \left( \ln { 2 } \right) }^{ { 2 }^{ -n } }$$

- 2 years, 11 months ago

Solve for n=1,2,3..find a pattern..then use principle of mathematical induction to find a recursion formula..

- 2 years, 11 months ago

- 2 years, 11 months ago

Ok I have done it probably..will need some time to write it down using the equation editor..requires Fubini's theorem..

- 2 years, 11 months ago

Actually the dots are a bit confusing..even if I think of n operations..then from where and how to begin this operations is ambiguous..

- 2 years, 11 months ago

- 2 years, 11 months ago

Finally I have posted my solution..shown my working as you asked.. but it is so big that it might contain errors..I am not sure..

- 2 years, 11 months ago

Nope nope..I was just guessing then..this is a version of log trig integrals..I am right now working on the problem..for n=0..the integrand reduces to (ln cosx) and the RHS reduces to -ln 2..but we need to generalise..

- 2 years, 11 months ago

how many terms are there in the right hand side of the square root

- 2 years, 11 months ago

Yes. There are $$n$$ terms.

- 2 years, 11 months ago

I am guessing that there are $$n$$ terms.

Staff - 2 years, 11 months ago

Okay . But it's better to be sure by asking to the one who post this . @Surya Prakash who post this question ?

- 2 years, 11 months ago

Good job

- 2 years, 11 months ago

- 2 years, 11 months ago