Hyperbolic Integral

Calculus Level 5

0 ln x cosh 2 x d x = ln ( A π B ) C γ \large \int_0^\infty \dfrac{ \ln x}{\cosh^2 x} \, dx =\ln{ \left( \frac { A\pi }{ B } \right) } -C\gamma

The equation above holds true for positive integers A , B A, B and C C , with A , B A,B coprime. Find A + B + C A+B+C .

Notation : γ \gamma denote the Euler-Mascheroni constant .


The answer is 6.

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2 solutions

Aditya Kumar
Mar 10, 2016

0 log x cosh 2 x d x = lim a 0 lim b 2 a b ( 4 0 x a 1 1 + e b x d x ) = lim a 0 lim b 2 a b ( b a ( 2 3 a 4 ) Γ ( a ) ζ ( a ) ) = log ( π 4 ) γ \begin{aligned} \int_0^{\infty} \frac{\log x}{\cosh^2x}\ \mathrm{d}x &=\lim_{a\to0} \lim_{b\to2}\frac{\partial}{ \partial a \partial b}\left(-4\int_0^{ \infty}\frac{x^{a-1}}{1+e^{b x}} \ dx\right)\\&= \lim_{a\to0} \lim_{b\to2}\frac{\partial}{\partial a \partial b}\left(b^{-a} (2^{3-a}-4) \Gamma(a) \zeta(a)\right)\\&=\log\left(\frac{\pi}{4}\right)-\gamma\end{aligned}

Can you expand your second last line? I think there are some missing steps here... Thanks.

Pi Han Goh - 5 years, 2 months ago
Incredible Mind
Mar 16, 2016

I used the series expansions

Why don't u post it as a solution?

Aditya Kumar - 5 years, 3 months ago

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Bcoz I find it time consuming writing latex....if there was a option to upload solution as images I would have gladly did so

incredible mind - 5 years, 2 months ago

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yes you may post the pic of your solution.

Aditya Kumar - 5 years, 2 months ago

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