nice-calculus-prove-that-cos-pix-2-cosh-pix-dx-1-2-

Question Number 140831 by mnjuly1970 last updated on 13/May/21

\dots \dots n i c e \dots . c a l c u l u s \dots \dots

p r o v e t h a t ::

ξ := \int_{- \infty}^{\infty} \frac{c o s (π x^{2})}{c o s h (π x)} d x = \frac{1}{\sqrt{2}} \dots .

\dots \dots .

Answered by mathmax by abdo last updated on 13/May/21

Φ = \int_{- \infty}^{+ \infty} \frac{\cos (π x^{2})}{ch (π x)} dx \Rightarrow Φ =_{π x = t} 2 \int_{0}^{+ \infty} \frac{\cos (t . \frac{t}{π})}{ch (t)} \frac{dt}{π}

= \frac{2}{π} \int_{0}^{+ \infty} \frac{\cos (\frac{t^{2}}{π})}{ch (t)} dt \Rightarrow Φ = \frac{4}{π} \int_{0}^{+ \infty} \frac{\cos (\frac{t^{2}}{π})}{e^{t} + e^{- t}} dt

= \frac{4}{π} \int_{0}^{\infty} \frac{e^{- t} \cos (\frac{t^{2}}{π})}{1 + e^{- 2 t}} dt \Rightarrow \frac{π}{4} Φ = \int_{0}^{\infty} e^{- t} \cos (\frac{t^{2}}{π}) \sum_{n = 0}^{\infty} e^{- 2 nt} dt

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (2 n + 1) t} \cos (\frac{t^{2}}{π}) dt

=_{(2 n + 1) t = y} \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- y} \cos (\frac{y^{2}}{π {(2 n + 1)}^{2}}) \frac{dy}{(2 n + 1)}

= \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} Re (\int_{0}^{\infty} e^{- y - i \frac{y^{2}}{π {(2 n + 1)}^{2}}} dy) we have

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

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

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

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

=_{\frac{y}{\sqrt{π} (2 n + 1)} - \sqrt{π} (2 n + 1) = z} e^{i π {(2 n +)}^{2}} \int_{- \sqrt{π} (2 n + 1)}^{+ \infty} e^{- {iz}^{2}} \sqrt{π} (2 n + 1) dz

\dots . be continued \dots