Hyperbolic Hyperdrive!

Calculus Level 4

Evaluate:

I = 0 e x 1 2 cosh ( x ) 1 d x \displaystyle I = \int_0^\infty \frac{\sqrt{e^{x} - 1}}{2\text{cosh}(x) - 1} \, dx

If I I can be expressed as π a b \frac{\pi^{a}}{\sqrt{b}} , where a , b N a,b \in \mathbb{N} and b b is square-free, then enter your answer as a + b a + b .


The answer is 4.

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1 solution

0 e x 1 2 c o s h x 1 d x = 2 0 t 2 t 4 + t 2 + 1 d t e x = 1 + t 2 = 2 0 d u u 4 + u 2 + 1 t 1 u = 0 x 1 / 2 x 2 + x + 1 d x u 2 x = π 3 \displaystyle \begin{aligned} \int_0^\infty \dfrac{\sqrt{e^x-1}}{2{\rm coshx}-1}\; dx &= 2\int_0^\infty \dfrac{t^2}{t^4+t^2+1}\; dt\quad \color{#3D99F6}{e^x=1+t^2} \\ &= 2\int_0^\infty \dfrac{du}{u^4+u^2+1}\quad \color{#3D99F6}{t\to \dfrac{1}{u}} \\ &= \int_0^\infty \dfrac{x^{-1/2}}{x^2+x+1}\; dx \quad \color{#3D99F6}{u^2\to x} \\&= \dfrac{\pi}{\sqrt{3}}\end{aligned}

In the last line we have used F ( a , s ) = 0 x s x 2 + 2 a x + 1 d x = π sin ( ( 1 s ) cos 1 a ) sin ( π s ) 1 a 2 \displaystyle F(a,s)=\int_{0}^{\infty} \frac{x^s}{x^2+2ax+1}dx = \frac{\pi\sin((1-s)\cos^{-1}a)}{\sin(\pi |s|)\sqrt{1-a^2}} which I have proved here

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