Calculus Challenge - 1 (Reposted)

Rn+:=2π∫0π/2x2+ln⁡2 ⁣cos⁡x2n12+1212+⋯+1212+12ln⁡2 ⁣cos⁡xx2+ln⁡2 ⁣cos⁡x dx\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 RnR_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.
#Calculus #SuryaPrakash #CalculusChallenges #Chellenges #KartikSharma

Note by Surya Prakash
5 years, 5 months ago

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Comments

In this case..first we observe that the nested radicals are of the form:12+a2\sqrt { \frac { 1 }{ 2 } +\frac { a }{ 2 } } which suggests that aa can be considered as cos⁡θ\cos { \theta } for some θ\theta ..Thus the nested radical part of the integral reduces to 12+1212+1212+...+1212+cos⁡θ2=cos⁡(θ2n)\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⁡θ=(ln⁡cos⁡x)2x2+(ln⁡cos⁡x)2⇒sec⁡2θ=x2+(ln⁡cos⁡x)2(ln⁡cos⁡x)2=1+x2(ln⁡cos⁡x)2=1+tan⁡2θ⇒tan⁡θ=±xln⁡cos⁡x\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 θ\theta to be such that tan⁡θ=−xln⁡cos⁡x\tan { \theta } =- \frac { x }{ \ln { \cos { x } } }. Now, let r∈(0,1)r\in \left( 0,1 \right) . We use the two following results:∫0+∞yr−1cos⁡(ay)e−bydy=Γ(r)cos⁡{rtan⁡−1(ab)}(a2+b2)r2\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 ∫0π2cos⁡(yx)cos⁡yxdx=π2y+1\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 ∫0π2cos⁡{rtan⁡−1(−xln⁡cos⁡x)}{x2+(ln⁡x)2}r2dx=1Γ(r)∫0π2{∫0+∞yr−1cos⁡(xy)eyln⁡cos⁡xdy}dx=1Γ(r)∫0π2∫0+∞yr−1cos⁡(xy)eyln⁡cos⁡xdydx=1Γ(r)∫0+∞yr−1{∫0π2cos⁡(xy)eyln⁡cos⁡xdx}dy=1Γ(r)∫0+∞yr−1{∫0π2cos⁡(xy)cos⁡yxdx}dy=1Γ(r)∫0+∞yr−1π2y+1dy=π2Γ(r)∫0+∞yr−12−ydy=π2Γ(r)∫0+∞yr−1e−yln⁡2dy=π2Γ(r).Γ(r)(ln⁡2)r=π2(ln⁡2)r\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>0r>0.)

We can apply analytic continuation to the above result to extend the domain of rr from (0,1)(0,1) to (−1,1)(-1,1) since the original domain (0,1)(0,1) is open in RR and the integrand function is analytic in rr. Now we simply put r=−12nr=-\frac { 1 }{ { 2 }^{ n } } to get that Rn+=2π.π2(ln⁡2)−12n=(ln⁡2)2−n{ 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 - 5 years, 5 months ago

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

Aman Rajput - 5 years, 5 months ago

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I am guessing that there are nn terms.

Calvin Lin Staff - 5 years, 5 months ago

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

Aman Rajput - 5 years, 5 months ago

Yes. There are nn terms.

Surya Prakash - 5 years, 5 months ago

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

Kuldeep Guha Mazumder - 5 years, 5 months ago

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This is easier said that done. Please show your working. Thanks.

Pi Han Goh - 5 years, 5 months ago

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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 - 5 years, 5 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..

Kuldeep Guha Mazumder - 5 years, 5 months ago

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@Kuldeep Guha Mazumder – @Surya Prakash please answer Kuldeep.

Pi Han Goh - 5 years, 5 months 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 - 5 years, 5 months ago

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 - 5 years, 5 months ago

Good job

Saleem Mansoor - 5 years, 5 months ago
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