CALCULUS-cos-pix-2-2-dx-

Question Number 127432 by mnjuly1970 last updated on 29/Dec/20

\dots CALCULUS \dots

\emptyset = \int_{- \infty}^{+ \infty} c o s (\frac{π x^{2}}{2}) d x = ?

Answered by Dwaipayan Shikari last updated on 29/Dec/20

\int_{- \infty}^{\infty} \sqrt{\frac{2}{π}} c o s (u^{2}) d u \sqrt{\frac{π}{2}} x = u \Rightarrow

= \sqrt{\frac{8}{π}} \int_{0}^{\infty} c o s (u^{2}) d u = 1

\int_{0}^{\infty} c o s (u^{2}) d u = \frac{1}{2} \int_{0}^{\infty} e^{u^{2} i} + e^{- u^{2} i} d u = \frac{1}{4} \int_{0}^{\infty} t^{- \frac{1}{2}} e^{i t} + t^{- \frac{1}{2}} e^{- i t} d t

= \frac{{(i)}^{- \frac{1}{2}} i}{4} \int_{0}^{\infty} m^{- \frac{1}{2}} e^{- m} d m - \frac{{(- i)}^{- \frac{1}{2}} i}{4} \int_{0}^{\infty} j^{- \frac{1}{2}} e^{- j} d j t = - i j t = i m

= 2 i Γ (\frac{1}{2}) (\frac{e^{- \frac{π}{4} i} - e^{\frac{π}{4} i}}{4}) = 2 \sqrt{\frac{π}{16}} s i n (\frac{π}{4}) = \sqrt{\frac{π}{8}}

Commented by mnjuly1970 last updated on 29/Dec/20

t h a n k y o u m a s t e r \dots

Commented by Dwaipayan Shikari last updated on 31/Dec/20

H a v e a g r e a t y e a r s i r!