0-sin-2-x-x-x-dx-pi-

Question Number 150828 by mnjuly1970 last updated on 15/Aug/21

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

Answered by puissant last updated on 16/Aug/21

Q = \int_{0}^{\infty} \frac{{s i n}^{2} x}{x \sqrt{x}} d x

= \int_{0}^{\infty} x^{- \frac{3}{2}} {s i n}^{2} x d x

= {[\frac{1}{1 - \frac{3}{2}} x^{- \frac{1}{2}} {s i n}^{2} x]}_{0}^{\infty} - \int_{0}^{\infty} - 2 x^{- \frac{1}{2}} \times 2 s i n x c o s x d x

= 2 \int_{0}^{\infty} \frac{s i n 2 x}{\sqrt{x}} d x

\underset{\sqrt{x} \to u}{=} 2 \int_{0}^{\infty} \frac{s i n 2 u^{2}}{u} (2 u d u)

= 4 \int_{0}^{\infty} s i n 2 u^{2} d u = - 4 i m (\int_{0}^{\infty} e^{- 2 {i u}^{2}} d u)

∵ ∵ ∵ \int_{0}^{\infty} e^{- {(\sqrt{2 i} u)}^{2}} d u \underset{\sqrt{2 i} u \to z}{=} \int_{0}^{z} e^{- z^{2}} \frac{d z}{\sqrt{2 i}}

= \frac{1}{\sqrt{2}} e^{- i \frac{π}{4}} \times \frac{\sqrt{π}}{2}

= \frac{\sqrt{π}}{2 \sqrt{2}} (\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) = \frac{\sqrt{π}}{4} - i \frac{\sqrt{π}}{4}

Q = - 4 i m (\int_{0}^{\infty} e^{- 2 {i u}^{2}} d u) = - 4 (- \frac{\sqrt{π}}{4})

∵∴ Q = \sqrt{π} . .

\dots \dots \dots \dots L e p u i s s a n t \dots \dots \dots .

Commented by mnjuly1970 last updated on 16/Aug/21

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

Answered by mathmax by abdo last updated on 17/Aug/21

Υ = \int_{0}^{\infty} \frac{\sin^{2} (x)}{x \sqrt{x}} dx =_{\sqrt{x} = t} \int_{0}^{\infty} \frac{\sin^{2} (t^{2})}{t^{3}} (2 t) dt

= 2 \int_{0}^{\infty} \frac{\sin^{2} (t^{2})}{t^{2}} dt = 2 {{[- \frac{1}{t} \sin^{2} (t^{2})]}_{0}^{\infty} - \int_{0}^{\infty} - \frac{1}{t} \times 2 \sin (t^{2}) 2 t) \cos (t^{2}) dt}

= 2 \int_{0}^{\infty} \sin ({2 t}^{2}) dt = 2 \int_{- \infty}^{+ \infty} \sin ({2 t}^{2}) dt = - 2 Im (\int_{- \infty}^{+ \infty} e^{- {2 it}^{2}} dt)

\int_{- \infty}^{+ \infty} e^{- (\sqrt{2} i) t^{2}} dt =_{\sqrt{2} it = y} \int_{- \infty}^{+ \infty} e^{- y^{2}} \frac{dy}{\sqrt{2} i}

= \frac{\sqrt{π}}{\sqrt{2} e^{\frac{i π}{4}}} = \frac{\sqrt{π}}{\sqrt{2}} e^{- \frac{i π}{4}} = \frac{\sqrt{π}}{\sqrt{2}} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{\sqrt{π}}{2} - \frac{i \sqrt{π}}{2} \Rightarrow

Υ = 2 \times \frac{\sqrt{π}}{2} = \sqrt{π}