Calculus Challenge 8!

Rn=2π0π/2(θ2+ln2(cosθ))2n112+1212++12ln2(cosθ)θ2+ln2(cosθ) dθ=(ln2)2n\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{-{2}^{-n}} Rn+=2π0π/2(θ2+ln2(cosθ))2n112+1212++12ln2(cosθ)θ2+ln2(cosθ) dθ=(ln2)2n \displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{{2}^{-n}}

Ramanujan Trigonometric Log Integrals ! (Ramanujan Again! I love him)

Prove the above 2 AMAZING integrals (I must add that adjective, pardon me!)

Note by Kartik Sharma
3 years, 11 months ago

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asd

John John - 2 years, 9 months ago

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@Ishan Singh

Kartik Sharma - 3 years, 11 months ago

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Can u post the solution?

Aditya Kumar - 3 years, 10 months ago

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Hint: The infinite surds reminds you of what? cos(x)=12+cos(2x)2\cos(x) = \sqrt{\frac{1}{2} + \frac{\cos(2x)}{2}}

Kartik Sharma - 3 years, 10 months ago

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@Kartik Sharma We can simplify the integral to Rn=2π0π/2(θ2+ln2(cosθ))2n1cos(12ncos1(ln2(cosθ)θ2+ln2(cosθ)))\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right) and

Rn+=2π0π/2(θ2+ln2(cosθ))2n1cos(12ncos1(ln2(cosθ)θ2+ln2(cosθ)))\displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right)

but what to do after that?

Samuel Jones - 3 years, 10 months ago

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