... nice calculus... prove that:: 𝛗=∫_(−∞) ^( +∞) (( cosh(px))/(cosh(x))) = (π/(cos(((πp)/2))))


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Question Number 133334 by mnjuly1970 last updated on 21/Feb/21

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $ϕ = \int_{- \infty}^{+ \infty} \frac{c o s h (p x)}{c o s h (x)} = \frac{π}{c o s (\frac{π p}{2})}$
	Answered by mnjuly1970 last updated on 21/Feb/21

	Commented by Dwaipayan Shikari last updated on 21/Feb/21

	${H a v e n}^{'} t t h o u g h t o f . G r e a t w a y s i r!$
	Commented by mnjuly1970 last updated on 21/Feb/21

	$s i n c e r e l y y o u r s . .$ $g r a t e f u l s i r p a y a n . . .$
	Answered by mathmax by abdo last updated on 22/Feb/21

	$Φ = \int_{- \infty}^{+ \infty} \frac{ch (px)}{ch (x)} dx \Rightarrow Φ = \int_{- \infty}^{+ \infty} \frac{e^{px} + e^{- px}}{e^{x} + e^{- x}} dx$ $=_{e^{x} = t} \int_{0}^{\infty} \frac{t^{p} + t^{- p}}{t + t^{- 1}} \frac{dt}{t} = \int_{0}^{\infty} \frac{t^{p} + t^{- p}}{t^{2} + 1} dt = \int_{0}^{\infty} \frac{t^{p}}{1 + t^{2}} dt + \int_{0}^{\infty} \frac{t^{- p}}{1 + t^{2}} dt$ $we have \int_{0}^{\infty} \frac{t^{p}}{1 + t^{2}} dt =_{t = z^{\frac{1}{2}}} \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{p}{2}}}{1 + z} z^{- \frac{1}{2}} dz$ $= \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{p + 1}{2} - 1}}{1 + z} dz = \frac{1}{2} \frac{π}{\sin (π (\frac{p + 1}{2}))} = \frac{π}{2 \cos (\frac{p π}{2})} also$ $\int_{0}^{\infty} \frac{t^{- p}}{1 + t^{2}} dt = \frac{π}{2 \cos (\frac{- p π}{2})} = \frac{π}{2 \cos (\frac{p π}{2})} \Rightarrow$ $Φ = \frac{π}{2 \cos (\frac{p π}{2})} + \frac{π}{2 \cos (\frac{p π}{2})} = \frac{π}{\cos (\frac{p π}{2})}$
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