Question-156915

Question Number 156915 by mnjuly1970 last updated on 17/Oct/21

Answered by qaz last updated on 07/Nov/21

\int_{0}^{\infty} \frac{dx}{(π^{2} + x^{2}) \cosh x}

= \int_{- \infty}^{+ \infty} \frac{e^{x}}{(e^{2 x} + 1) (π^{2} + x^{2})} dx

= \int_{- \infty}^{+ \infty} \frac{e^{x}}{e^{2 x} + 1} \cdot \frac{1}{π} ℜ \int_{0}^{\infty} e^{- (π - ix) y} dydx

= \frac{1}{π} ℜ \int_{0}^{\infty} e^{- π y} dy \int_{- \infty}^{+ \infty} \frac{e^{x (1 + iy)}}{e^{2 x} + 1} dx

= \frac{1}{2 π} ℜ \int_{0}^{\infty} e^{- π y} dy \int_{0}^{\infty} \frac{t^{\frac{iy - 1}{2}}}{t + 1} dt \dots \dots . e^{x} \to t

= \frac{1}{2} ℜ \int_{0}^{\infty} \frac{e^{- π y}}{\sin (\frac{iy + 1}{2} π)} dy

= \frac{1}{2} \int_{0}^{\infty} \frac{e^{- π y}}{\cosh \frac{π y}{2}} dy

= \frac{1}{π} \int_{0}^{\infty} \frac{{2 e}^{- 3 y}}{1 + e^{- 2 y}} dy

= \frac{2}{π} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 3) y} dy

= \frac{2}{π} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 3}

= \frac{2}{π} (1 - \frac{π}{4})

= \frac{2}{π} - \frac{1}{2}

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