Question-35544

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))

$${let}\:\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{cosx}}{\:\sqrt{{x}}}{dx}\:\:{and}\:{B}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{\:\sqrt{{x}}}{dx}\:{we}\:{have} \\ $$$${A}\:−{iB}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}\:\:{changement}\:\sqrt{{x}}\:={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{it}^{\mathrm{2}} } }{{t}}\mathrm{2}{tdt} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{it}^{\mathrm{2}} } {dt}\:\:\:=_{{t}\sqrt{{i}}=\:{u}} \:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{{i}}} \\ $$$$=\:\frac{\mathrm{2}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\:\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$=\sqrt{\pi}\left\{\:{cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosx}}{\:\sqrt{{x}}}{dx}\:=\sqrt{\pi}\:\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{\:\sqrt{{x}}}{dx}\:=\sqrt{\pi}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$

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