prove-that-0-1-2-x-sin-pi-2-x-1-x-dx-1-8pi-m-n-1970-

Question Number 117380 by mnjuly1970 last updated on 11/Oct/20

\dots p r o v e t h a t \dots

Ω = \int_{0}^{\infty} \frac{1}{2 \sqrt{x}} s i n (π^{2} x + \frac{1}{x}) d x = \frac{1}{\sqrt{8 π}}

m . n . 1970

Commented by mindispower last updated on 11/Oct/20

n i c e o n e

Answered by mnjuly1970 last updated on 13/Oct/20

s o l u t i o n ::

R e c a l l :: \int_{0}^{\infty} s i n (z^{2}) d z \overset{. . f r e s n e l i n t e g r a l . .}{=} \sqrt{\frac{π}{8}}

Ω \overset{t = \sqrt{x}}{=} \int_{0}^{\infty} s i n (π^{2} t^{2} + \frac{1}{t^{2}}) d t

π Ω = \int_{0}^{\infty} π s i n (π^{2} t^{2} + \frac{1}{t^{2}}) d t (i)

π Ω \overset{t = \frac{1}{π u}}{=} \frac{1}{π} \int_{0}^{\infty} π s i n (\frac{1}{u^{2}} + π^{2} u^{2}) \frac{d u}{u^{2}}

π Ω = \int_{0}^{\infty} s i n (π^{2} u^{2} + \frac{1}{u^{2}}) \frac{d u}{u^{2}} (i i)

(i) + (i i) ::

2 π Ω = \int_{0}^{\infty} (π + \frac{1}{x^{2}}) s i n [{(π x - \frac{1}{x})}^{2} + 2 π] d u

2 π Ω \overset{π x - \frac{1}{x} = y}{=} \int_{- \infty}^{\infty} s i n (y^{2}) d y

2 π Ω \overset{R e c a l l}{=} 2 \sqrt{\frac{π}{8}} \Rightarrow Ω = \sqrt{\frac{1}{8 π}} ✓

\dots ♣ M . N . j u l y . 1970 ♣ \dots

♠ p e a c e b e u p o n y o u ♠