advanced-calculus-prove-that-0-x-2-cosh-2-x-2-dx-2-2-4-pi-1-2-

Question Number 140715 by mnjuly1970 last updated on 11/May/21

\dots \dots a d v a n c e d c a l c u l u s \dots \dots

p r o v e t h a t :

ϕ := \int_{0}^{\infty} \frac{x^{2}}{{c o s h}^{2} (x^{2})} d x \overset{?}{=} \frac{\sqrt{2} - 2}{4} \sqrt{π} ζ (\frac{1}{2})

\dots \dots \dots \dots . .

Answered by Dwaipayan Shikari last updated on 11/May/21

\int_{0}^{\infty} \frac{x^{2}}{{c o s h}^{2} (x^{2})} d x = \frac{1}{2} \int_{0}^{\infty} \frac{\sqrt{u}}{{c o s h}^{2} (u)} d u

= \int_{0}^{\infty} \frac{\sqrt{u}}{1 + c o s h (2 u)} = 2 \int_{0}^{\infty} \frac{e^{- 2 u} \sqrt{u}}{{(1 + e^{- 2 u})}^{2}} d u

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} e^{- 2 n u} = \frac{1}{1 + e^{- 2 u}} \Rightarrow \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} {n e}^{- 2 n u} = \frac{- e^{2 u}}{{(1 + e^{- 2 u})}^{2}}

= - 2 \int_{0}^{\infty} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \sqrt{u} {n e}^{- 2 n u} d u = 2 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} n \int_{0}^{\infty} \sqrt{u} e^{- 2 n u} d u

2 n u = t \Rightarrow \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{\infty} \frac{\sqrt{t}}{\sqrt{2 n}} e^{- t} d t = - \underset{n = 1}{\sum^{\infty}} \frac{\sqrt{π}}{\sqrt{2} \sqrt{n}} {(- 1)}^{n + 1}

= - \sqrt{\frac{π}{2}} ζ (\frac{1}{2}) (1 - \frac{1}{2^{\frac{1}{2} - 1}}) = \sqrt{\frac{π}{2}} (\sqrt{2} - 1) ζ (\frac{1}{2})

Commented by mnjuly1970 last updated on 12/May/21

g r a t e f u l \dots .