.....nice calculus..... prove that::: 𝛗=∫_0 ^( ∞) ((cos(2x))/(cosh(x))) dx=^? (π/(2cosh(π)))


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

$. . . . . n i c e c a l c u l u s . . . . .$ $p r o v e t h a t :::$ $ϕ = \int_{0}^{\infty} \frac{c o s (2 x)}{c o s h (x)} d x \overset{?}{=} \frac{π}{2 c o s h (π)}$
	Answered by mindispower last updated on 10/Feb/21
	$(1/(ch(x)))=((2e^x )/(e^(2x) +1))=2e^(−x) Σ_(k≥0) (−1)^k e^(−2kx) =Σ_(k≥0) (−1)^k e^(−x(2k+1)) ....true ∀x>0 φ=(1/2)Re∫_(−∞) ^∞ 2((e^(2ix+x) )/(e^(2x) +1))dx let C ={x,Imx≥0} e^(2x) +1=0⇒2x=iπ+2ikπ x_k =i((π/2)+kπ),k≥0 Res((e^(2ix+x) /(e^(2x) +1)),x_k )=((e^(−π−2kπ) .i.(−1)^k )/(2(−1))) ∫_C (e^(2ix+x) /(e^(2x) +1))dx=2iπRes((e^(2ix+x) /(e^(2x) +1)),x∈C) =2iπΣ_(k≥0) ((e^(−π−2kπ) i(−1)^k )/(−2)) =πe^(−π) Σ_(k≥0) (−e^(−2π) )^k =((πe^(−π) )/(1+e^(−2π) ))=(π/(e^π +e^(−π) ))=(π/(2ch(π))) we get so ∫_0 ^∞ ((cos(2x))/(ch(x)))dx=(1/2).Re.2(π/(2ch(π)))=(π/(2cosh(π)))$
	$\frac{1}{c h (x)} = \frac{2 e^{x}}{e^{2 x} + 1} = 2 e^{- x} \sum_{k ⩾ 0} {(- 1)}^{k} e^{- 2 k x}$ $= \sum_{k ⩾ 0} {(- 1)}^{k} e^{- x (2 k + 1)} . . . . t r u e \forall x > 0$ $ϕ = \frac{1}{2} R e \int_{- \infty}^{\infty} 2 \frac{e^{2 i x + x}}{e^{2 x} + 1} d x$ $l e t C = {x, I m x ⩾ 0}$ $e^{2 x} + 1 = 0 \Rightarrow 2 x = i π + 2 i k π$ $x_{k} = i (\frac{π}{2} + k π), k ⩾ 0$ $R e s (\frac{e^{2 i x + x}}{e^{2 x} + 1}, x_{k}) = \frac{e^{- π - 2 k π} . i . {(- 1)}^{k}}{2 (- 1)}$ $\int_{C} \frac{e^{2 i x + x}}{e^{2 x} + 1} d x = 2 i π R e s (\frac{e^{2 i x + x}}{e^{2 x} + 1}, x \in C)$ $= 2 i π \sum_{k ⩾ 0} \frac{e^{- π - 2 k π} i {(- 1)}^{k}}{- 2}$ $= π e^{- π} \sum_{k ⩾ 0} {(- e^{- 2 π})}^{k} = \frac{π e^{- π}}{1 + e^{- 2 π}} = \frac{π}{e^{π} + e^{- π}} = \frac{π}{2 c h (π)}$ $w e g e t s o$ $\int_{0}^{\infty} \frac{c o s (2 x)}{c h (x)} d x = \frac{1}{2} . R e . 2 \frac{π}{2 c h (π)} = \frac{π}{2 c o s h (π)}$
	Commented by mnjuly1970 last updated on 10/Feb/21

	$e x c e l l e n t . .$ $t h a n k s a l o t s i r p o w e r . . .$
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