0-cos-pix-pi-2-x-2-dx-

Question Number 85346 by naka3546 last updated on 21/Mar/20

\int_{0} \overset{\infty}{} \frac{\cos (π x)}{π^{2} + x^{2}} d x = ?

Commented by abdomathmax last updated on 21/Mar/20

A = \int_{0}^{\infty} \frac{c o s (π x)}{x^{2} + π^{2}} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{c o s (π x)}{x^{2} + π^{2}} d x

= R e (\int_{- \infty}^{+ \infty} \frac{e^{i π x}}{x^{2} + π^{2}} d x) l e t φ (z) = \frac{e^{i π z}}{z^{2} + π^{2}}

w e h a v e φ (z) = \frac{e^{i π z}}{(z - i π) (z + i π)}

r e s i d u s t h e i r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i π)

= 2 i π \frac{e^{i π (i π)}}{2 i π} = e^{- π^{2}} \Rightarrow A = \frac{e^{- π^{2}}}{2}

Answered by M±th+et£s last updated on 21/Mar/20

f (p) = \int_{0}^{\infty} \frac{c o s (p π x)}{π^{2} + x^{2}} d x \Rightarrow\Rightarrow\Rightarrow ⌊ (f (p)) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{c o s (p π x)}{π^{2} + x^{2}} d x d p

= \int_{0}^{\infty} \frac{1}{π^{2} + x^{2}} \int_{0}^{\infty} e^{- p s} c o s (p π x) d x d p

= \int_{0}^{\infty} \frac{1}{(π^{2} + x^{2})} (\frac{s}{s^{2} + π^{2} x^{2}}) d x

= \frac{s}{s - π^{4}} \int_{0}^{\infty} \frac{s^{2} + π^{2} x^{2} - π^{4} - π^{2} x^{2}}{(π^{2} + x^{2}) (s^{2} + π^{2} x^{2})} d x

\frac{s}{s^{2} - π^{4}} \int (\frac{1}{π^{2} + x^{2}} - \frac{π^{2}}{s^{2} + π^{2} x^{2}}) d x

\frac{s}{s^{2} - π^{4}} {[\frac{1}{π} {t a n}^{- 1} \frac{x}{π} - \frac{π}{s} {t a n}^{- 1} \frac{π x}{s}]}_{0}^{\infty}

\frac{s}{s^{2} - π^{4}} [\frac{1}{2} - \frac{π^{2}}{2 s}]

\frac{s}{s^{2} - π^{4}} (\frac{s - π^{2}}{2 s}) = \frac{1}{2 (s + π^{2})}

f (p) = ⌊^{- 1} (\frac{1}{s + π^{2}})

= \frac{1}{2} e^{- π^{2}}