nice..mathematics... Prove that... I= ∫_0 ^( ∞) (( cos (x ))/(cosh (x ))) dx=(π/( cosh ((π/2) ))) .......■ prepared


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Question Number 152797 by mnjuly1970 last updated on 01/Sep/21

$n i c e . . m a t h e m a t i c s . . .$ $Prove that . . .$ $I = \int_{0}^{\infty} \frac{c o s (x)}{c o s h (x)} d x = \frac{π}{c o s h (\frac{π}{2})} . . . . . . . ◼$ $p r e p a r e d :: m . n$
	Answered by mnjuly1970 last updated on 01/Sep/21
	$I:= ∫_0 ^( ∞) (( e^( ix) +e^( −ix) )/(e^( x) +e^( −x) ))dx= ∫_0 ^( ∞) (( e^( x(i−1)) +e^( −x(i+1)) )/(1+e^( −2x) ))dx :=^(e^( −2x) =y) (1/2)∫_0 ^( 1) (( y^( ((1−i)/2)) +y^((1+i)/2) )/(1+ y)) (dy/y) := (1/2) ∫_0 ^( 1) (( y^( −((1+i)/2)) +y^( −((1−i)/2)) )/(1+y)) dy := (1/2) ∫_0 ^( 1) (( (1−y )( y^( −((1+i)/2)) +y^(−((1−i)/2)) ))/(1−y^( 2) ))dy := (1/2) ∫_0 ^( 1) (( y^( −((1+i)/2)) +y^(−((1−i)/2)) −y^((1−i)/2) −y^( ((1+i)/2)) )/(1−y^( 2) ))dy :=^(y^( 2) = t) (1/4) ∫_0 ^( 1) (( t^( ((−3−i)/4)) −1+t^((−3+i)/4) −1+1−t^((−1−i)/4) +1−t^( ((−1+i)/4)) )/(1−t))dt := (1/4) (ψ (((3−i)/4) )+[ψ(((3+i)/4))−ψ(((1−i)/4))]−ψ(((1+i)/4))) :=^(ψ (z )−ψ (1−z )= −π cot(πz )) (1/4) (−πcot ((((3+i)π)/4))−πcot((((3−i)π)/4) )) :=(π/4) {cot((π/4) −((iπ)/4) )+ cot((π/4) +((iπ)/4))} := (π/4) {(( 1+tan(((iπ)/4)))/(1−tan(((iπ)/4) ))) +((1−tan(((iπ)/4)))/(1+tan(((iπ)/4)))) } := (π/4) { (( 2( 1+tan^( 2) (((iπ)/4) )))/(1−tan^( 2) (((iπ)/4))))} := (π/2) ((( 1)/(cos (i(π/2)))))= (π/(2cosh ((π/2) ))) ...■ m.n$
	$I := \int_{0}^{\infty} \frac{e^{i x} + e^{- i x}}{e^{x} + e^{- x}} d x = \int_{0}^{\infty} \frac{e^{x (i - 1)} + e^{- x (i + 1)}}{1 + e^{- 2 x}} d x$ $: \overset{e^{- 2 x} = y}{=} \frac{1}{2} \int_{0}^{1} \frac{y^{\frac{1 - i}{2}} + y^{\frac{1 + i}{2}}}{1 + y} \frac{d y}{y}$ $:= \frac{1}{2} \int_{0}^{1} \frac{y^{- \frac{1 + i}{2}} + y^{- \frac{1 - i}{2}}}{1 + y} d y$ $:= \frac{1}{2} \int_{0}^{1} \frac{(1 - y) (y^{- \frac{1 + i}{2}} + y^{- \frac{1 - i}{2}})}{1 - y^{2}} d y$ $:= \frac{1}{2} \int_{0}^{1} \frac{y^{- \frac{1 + i}{2}} + y^{- \frac{1 - i}{2}} - y^{\frac{1 - i}{2}} - y^{\frac{1 + i}{2}}}{1 - y^{2}} d y$ $: \overset{y^{2} = t}{=} \frac{1}{4} \int_{0}^{1} \frac{t^{\frac{- 3 - i}{4}} - 1 + t^{\frac{- 3 + i}{4}} - 1 + 1 - t^{\frac{- 1 - i}{4}} + 1 - t^{\frac{- 1 + i}{4}}}{1 - t} d t$ $:= \frac{1}{4} (ψ (\frac{3 - i}{4}) + [ψ (\frac{3 + i}{4}) - ψ (\frac{1 - i}{4})] - ψ (\frac{1 + i}{4}))$ $: \overset{ψ (z) - ψ (1 - z) = - π c o t (π z)}{=} \frac{1}{4} (- π c o t (\frac{(3 + i) π}{4}) - π c o t (\frac{(3 - i) π}{4}))$ $:= \frac{π}{4} {c o t (\frac{π}{4} - \frac{i π}{4}) + c o t (\frac{π}{4} + \frac{i π}{4})}$ $:= \frac{π}{4} {\frac{1 + t a n (\frac{i π}{4})}{1 - t a n (\frac{i π}{4})} + \frac{1 - t a n (\frac{i π}{4})}{1 + t a n (\frac{i π}{4})}}$ $:= \frac{π}{4} {\frac{2 (1 + {t a n}^{2} (\frac{i π}{4}))}{1 - {t a n}^{2} (\frac{i π}{4})}}$ $:= \frac{π}{2} (\frac{1}{c o s (i \frac{π}{2})}) = \frac{π}{2 c o s h (\frac{π}{2})} . . . ◼ m . n$
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