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Question Number 35544 by tanmay.chaudhury50@gmail.com last updated on 20/May/18

Commented by prof Abdo imad last updated on 20/May/18
$let A =∫_0 ^∞ ((cosx)/(√x))dx and B =∫_0 ^∞ ((sinx)/(√x))dx we have A −iB = ∫_0 ^∞ (e^(−ix) /(√x))dx changement (√x) =t give ∫_0 ^∞ (e^(−ix) /(√x))dx = ∫_0 ^∞ (e^(−it^2 ) /t)2tdt = 2 ∫_0 ^∞ e^(−it^2 ) dt =_(t(√i)= u) 2 ∫_0 ^∞ e^(−u^2 ) (dt/(√i)) = (2/e^(i(π/4)) ) ∫_0 ^∞ e^(−u^2 ) du = 2 e^(−i(π/4)) ((√π)/2) =(√π){ cos((π/4))−isin((π/4))} ⇒ ∫_0 ^∞ ((cosx)/(√x))dx =(√π) .(1/(√2)) =(√(π/2)) ∫_0 ^∞ ((sinx)/(√x))dx =(√π) (1/(√2)) =(√(π/2))$
$l e t A = \int_{0}^{\infty} \frac{c o s x}{\sqrt{x}} d x a n d B = \int_{0}^{\infty} \frac{s i n x}{\sqrt{x}} d x w e h a v e$ $A - i B = \int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t g i v e$ $\int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x = \int_{0}^{\infty} \frac{e^{- {i t}^{2}}}{t} 2 t d t$ $= 2 \int_{0}^{\infty} e^{- {i t}^{2}} d t =_{t \sqrt{i} = u} 2 \int_{0}^{\infty} e^{- u^{2}} \frac{d t}{\sqrt{i}}$ $= \frac{2}{e^{i \frac{π}{4}}} \int_{0}^{\infty} e^{- u^{2}} d u = 2 e^{- i \frac{π}{4}} \frac{\sqrt{π}}{2}$ $= \sqrt{π} {c o s (\frac{π}{4}) - i s i n (\frac{π}{4})} \Rightarrow$ $\int_{0}^{\infty} \frac{c o s x}{\sqrt{x}} d x = \sqrt{π} . \frac{1}{\sqrt{2}} = \sqrt{\frac{π}{2}}$ $\int_{0}^{\infty} \frac{s i n x}{\sqrt{x}} d x = \sqrt{π} \frac{1}{\sqrt{2}} = \sqrt{\frac{π}{2}}$
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