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Question Number 62262 by maxmathsup by imad last updated on 18/Jun/19
find the value of I =∫_0 ^∞ ((e^(−t) sint)/( (√t)))dt and J =∫_0 ^∞ ((e^(−t) cos(t))/( (√t)))dt ,study first the convergence.
$${find}\:{the}\:{value}\:{of}\: \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} {sint}}{\:\sqrt{{t}}}{dt}\:\:{and}\:{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} {cos}\left({t}\right)}{\:\sqrt{{t}}}{dt}\:\:,{study}\:{first}\:{the}\:{convergence}. \\ $$
Commented by maxmathsup by imad last updated on 19/Jun/19
the convergence o I and J is proved by sir smail let passe to the values we have J−iI =∫_0 ^∞ (e^(−t) /( (√t)))(cost −isint)dt =∫_0 ^∞ ((e^(−t ) e^(−it) )/( (√t))) dt =∫_0 ^∞ (e^(−(i+1)t) /( (√t))) dt changement (√t)=x give J−iI =∫_0 ^∞ (e^(−(1+i)x^2 ) /x) (2x)dx =2 ∫_0 ^∞ e^(−(1+i)x^2 ) dx =∫_(−∞) ^(+∞) e^(−(1+i)x^2 ) dx =_((√(1+i))x =u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/( (√(1+i)))) =(1/( (√(1+i)))) ∫_(−∞) ^(+∞) e^(−u^2 ) du =((√π)/( (√(1+i)))) 1+i =(√2)e^((iπ)/4) ⇒(√(1+i))=^4 (√2)e^((iπ)/8) ⇒J−iI =((√π)/((^4 (√2)))) e^(−((iπ)/8)) =((√π)/((^4 (√2)))){cos((π/8))−isin((π/8))} ⇒ I =((√π)/((^4 (√2)))) sin((π/8)) and J =((√π)/((^4 (√2)))) cos((π/8)) ⇒ I =((√π)/((^4 (√2))))((√(2−(√2)))/2) and J =((√π)/((^4 (√2)))) ((√(2+(√2)))/2) .
$${the}\:{convergence}\:{o}\:{I}\:{and}\:{J}\:{is}\:{proved}\:{by}\:{sir}\:{smail}\:{let}\:{passe}\:{to}\:{the}\:{values} \\ $$$${we}\:{have}\:{J}−{iI}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}\left({cost}\:−{isint}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}\:} {e}^{−{it}} }{\:\sqrt{{t}}}\:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({i}+\mathrm{1}\right){t}} }{\:\sqrt{{t}}}\:{dt}\:\:{changement}\:\sqrt{{t}}={x}\:{give} \\ $$$${J}−{iI}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−\left(\mathrm{1}+{i}\right){x}^{\mathrm{2}} } }{{x}}\:\left(\mathrm{2}{x}\right){dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{1}+{i}\right){x}^{\mathrm{2}} } {dx}\:=\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\mathrm{1}+{i}\right){x}^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{\mathrm{1}+{i}}{x}\:={u}} \:\:\:\:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{\mathrm{1}+{i}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{i}}}\:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{1}+{i}}} \\ $$$$\mathrm{1}+{i}\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\sqrt{\mathrm{1}+{i}}=^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\Rightarrow{J}−{iI}\:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \\ $$$$=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left\{{cos}\left(\frac{\pi}{\mathrm{8}}\right)−{isin}\left(\frac{\pi}{\mathrm{8}}\right)\right\}\:\Rightarrow\:{I}\:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\:{sin}\left(\frac{\pi}{\mathrm{8}}\right)\:{and}\:{J}\:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\:{cos}\left(\frac{\pi}{\mathrm{8}}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\:\:\:{and}\:\:{J}\:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\:\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\:. \\ $$
Answered by Smail last updated on 18/Jun/19
sint≤1⇔e^(−t) sint≤e^(−t) ((e^(−t) sint)/( (√t)))≤(e^(−t) /( (√t)))⇔∫_0 ^∞ ((e^(−t) sint)/( (√t)))dt≤∫_0 ^∞ (e^(−t) /( (√t)))dt Knowing that ∫_0 ^∞ (e^(−t) /( (√t)))dt=2∫_0 ^∞ e^(−x^2 ) dx=(√π) which means ∫_0 ^∞ (e^(−t) /( (√t)))dt converges. That means I=∫_0 ^∞ ((e^(−t) sint)/( (√t)))dt converges because I≤(√π) Also J≤∫_0 ^∞ (e^(−t) /( (√t)))dt meaning J also converges J−iI=∫_0 ^∞ ((e^(−t) e^(−it) )/( (√t)))dt =∫_0 ^∞ (e^(−t(1+i)) /( (√t)))dt Let x=(√t)((√(1+i)))⇒(dt/( (√t)))=((2dx)/( (√(1+i)))) J−iI=(2/( (2)^(1/4) e^(i(π/8)) ))∫_0 ^∞ e^(−x^2 ) dx=((√π)/( (2)^(1/4) e^(i(π/8)) )) =((√π)/( (2)^(1/4) ))(cos((π/8))−isin((π/8))) J=((√π)/( (2)^(1/4) ))cos((π/8))=((√π)/( (2)^(1/4) ))(√(((√2)+1)/(2(√2))))= J=((√(π((√2)+1)))/2) I=((√(π((√2)−1)))/2)
$${sint}\leqslant\mathrm{1}\Leftrightarrow{e}^{−{t}} {sint}\leqslant{e}^{−{t}} \\ $$$$\frac{{e}^{−{t}} {sint}}{\:\sqrt{{t}}}\leqslant\frac{{e}^{−{t}} }{\:\sqrt{{t}}}\Leftrightarrow\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} {sint}}{\:\sqrt{{t}}}{dt}\leqslant\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt} \\ $$$${Knowing}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi} \\ $$$${which}\:{means}\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}\:{converges}. \\ $$$${That}\:{means}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} {sint}}{\:\sqrt{{t}}}{dt}\:{converges}\: \\ $$$${because}\:{I}\leqslant\sqrt{\pi}\: \\ $$$${Also}\:\:{J}\leqslant\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}\:{meaning}\:\:{J}\:{also}\:{converges} \\ $$$${J}−{iI}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} {e}^{−{it}} }{\:\sqrt{{t}}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}\left(\mathrm{1}+{i}\right)} }{\:\sqrt{{t}}}{dt} \\ $$$${Let}\:\:{x}=\sqrt{{t}}\left(\sqrt{\mathrm{1}+{i}}\right)\Rightarrow\frac{{dt}}{\:\sqrt{{t}}}=\frac{\mathrm{2}{dx}}{\:\sqrt{\mathrm{1}+{i}}} \\ $$$${J}−{iI}=\frac{\mathrm{2}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{8}}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{8}}} } \\ $$$$=\frac{\sqrt{\pi}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\left({cos}\left(\frac{\pi}{\mathrm{8}}\right)−{isin}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$${J}=\frac{\sqrt{\pi}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}{cos}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\sqrt{\pi}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\sqrt{\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}= \\ $$$${J}=\frac{\sqrt{\pi\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}}{\mathrm{2}} \\ $$$${I}=\frac{\sqrt{\pi\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 18/Jun/19
thanks sir
$${thanks}\:{sir} \\ $$

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