Calculus Challenge 8!

\[\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{-{2}^{-n}} \] \[ \displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2}{{(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \cdots + \frac{1}{2}\sqrt{\frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)}}}} \ d\theta} = {(\ln 2)}^{{2}^{-n}}\]

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
2 years, 10 months ago

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asd

John John - 1 year, 8 months ago

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@Ishan Singh

Kartik Sharma - 2 years, 10 months ago

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Can u post the solution?

Aditya Kumar - 2 years, 9 months ago

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

Kartik Sharma - 2 years, 9 months ago

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@Kartik Sharma We can simplify the integral to \[\displaystyle R^-_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{-{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right) \] and

\[\displaystyle R^+_n = \frac{2}{\pi} \int_{0}^{\pi/2} {(\theta^2 + \ln^2(\cos \theta))}^{{2}^{-n-1}} \cdot \cos \left(\dfrac{1}{2^n} \cos^{-1} \left(\sqrt{ \frac{\ln^2(\cos \theta)}{\theta^2 + \ln^2(\cos \theta)} }\right) \right) \]

but what to do after that?

Samuel Jones - 2 years, 9 months ago

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