nice-mathematics-Prove-that-I-0-cos-x-cosh-x-dx-pi-cosh-pi-2-prepared-m-n-

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 \dots

Prove that \dots

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})} \dots \dots .

p r e p a r e d :: m . n

Answered by mnjuly1970 last updated on 01/Sep/21

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})} \dots m . n