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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
11 months, 3 weeks 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 \({ R }_{ n }^{ + }=\frac { 2 }{ \pi } .\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ -\frac { 1 }{ { 2 }^{ n } } } } ={ \left( \ln { 2 } \right) }^{ { 2 }^{ -n } }\) Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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Solve for n=1,2,3..find a pattern..then use principle of mathematical induction to find a recursion formula.. Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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@Kuldeep Guha Mazumder This is easier said that done. Please show your working. Thanks. Pi Han Goh · 11 months, 2 weeks ago

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@Pi Han Goh Ok I have done it probably..will need some time to write it down using the equation editor..requires Fubini's theorem.. Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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@Pi Han Goh 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.. Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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@Kuldeep Guha Mazumder @Surya Prakash please answer Kuldeep. Pi Han Goh · 11 months, 2 weeks ago

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@Pi Han Goh 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.. Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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@Pi Han Goh 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.. Kuldeep Guha Mazumder · 11 months, 2 weeks ago

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how many terms are there in the right hand side of the square root Aman Rajput · 11 months, 2 weeks ago

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@Aman Rajput Yes. There are \(n\) terms. Surya Prakash · 11 months, 2 weeks ago

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@Aman Rajput I am guessing that there are \(n\) terms. Calvin Lin Staff · 11 months, 2 weeks ago

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@Calvin Lin Okay . But it's better to be sure by asking to the one who post this . @Surya Prakash who post this question ? Aman Rajput · 11 months, 2 weeks ago

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Good job Saleem Mansoor · 11 months ago

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@Pi Han Goh @Aditya Kumar @Julian Poon Surya Prakash · 11 months, 2 weeks ago

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