nice-calculus-prove-that-0-cos-2x-cosh-x-dx-pi-2cosh-pi-

Question Number 132023 by mnjuly1970 last updated on 10/Feb/21

\dots . . n i c e c a l c u l u s \dots . .

p r o v e t h a t :::

ϕ = \int_{0}^{\infty} \frac{c o s (2 x)}{c o s h (x)} d x \overset{?}{=} \frac{π}{2 c o s h (π)}

Answered by mindispower last updated on 10/Feb/21

\frac{1}{c h (x)} = \frac{2 e^{x}}{e^{2 x} + 1} = 2 e^{- x} \sum_{k ⩾ 0} {(- 1)}^{k} e^{- 2 k x}

= \sum_{k ⩾ 0} {(- 1)}^{k} e^{- x (2 k + 1)} \dots . t r u e \forall x > 0

ϕ = \frac{1}{2} R e \int_{- \infty}^{\infty} 2 \frac{e^{2 i x + x}}{e^{2 x} + 1} d x

l e t C = {x, I m x ⩾ 0}

e^{2 x} + 1 = 0 \Rightarrow 2 x = i π + 2 i k π

x_{k} = i (\frac{π}{2} + k π), k ⩾ 0

R e s (\frac{e^{2 i x + x}}{e^{2 x} + 1}, x_{k}) = \frac{e^{- π - 2 k π} . i . {(- 1)}^{k}}{2 (- 1)}

\int_{C} \frac{e^{2 i x + x}}{e^{2 x} + 1} d x = 2 i π R e s (\frac{e^{2 i x + x}}{e^{2 x} + 1}, x \in C)

= 2 i π \sum_{k ⩾ 0} \frac{e^{- π - 2 k π} i {(- 1)}^{k}}{- 2}

= π e^{- π} \sum_{k ⩾ 0} {(- e^{- 2 π})}^{k} = \frac{π e^{- π}}{1 + e^{- 2 π}} = \frac{π}{e^{π} + e^{- π}} = \frac{π}{2 c h (π)}

w e g e t s o

\int_{0}^{\infty} \frac{c o s (2 x)}{c h (x)} d x = \frac{1}{2} . R e . 2 \frac{π}{2 c h (π)} = \frac{π}{2 c o s h (π)}

Commented by mnjuly1970 last updated on 10/Feb/21

e x c e l l e n t . .

t h a n k s a l o t s i r p o w e r \dots