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Question Number 35229 by abdo mathsup 649 cc last updated on 16/May/18
find the value of integral ∫_0 ^∞ e^(−(2+ia)^2 t^2 ) dt with a from R ∣a∣<1.
$${find}\:{the}\:{value}\:{of}\:{integral} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\mathrm{2}+{ia}\right)^{\mathrm{2}} {t}^{\mathrm{2}} } {dt}\:\:\:\:{with}\:{a}\:{from}\:{R}\:\:\:\:\mid{a}\mid<\mathrm{1}. \\ $$
Commented by abdo mathsup 649 cc last updated on 17/May/18
Remark we have A(α)= ∫_0 ^∞ e^(−(4 +4iα −α^2 )t^2 ) dt = ∫_0 ^∞ e^(−(4 −α^2 )t^2 −4iαt^2 ) dt = ∫_0 ^∞ e^((α^2 −4)t^2 ) {cos(4αt^2 ) −isin(4αt^2 )}dt = ∫_0 ^∞ e^((α^2 −4)t^2 ) cos(4αt^2 ) −i ∫_0 ^∞ e^((α^2 −4)t^2 ) sin(4αt^2 )dt ⇒ ∫_0 ^∞ e^((α^2 −4)t^2 ) cos(4αt^2 )dt = ((√π)/(4+α^2 )) and ∫_0 ^∞ e^((α^2 −4)t^2 ) sin(4αt^2 )dt = ((α(√π))/(8 +2α^2 )) .
$${Remark}\:\:{we}\:{have}\:{A}\left(\alpha\right)=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{4}\:+\mathrm{4}{i}\alpha\:−\alpha^{\mathrm{2}} \right){t}^{\mathrm{2}} } {dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−\left(\mathrm{4}\:−\alpha^{\mathrm{2}} \right){t}^{\mathrm{2}} −\mathrm{4}{i}\alpha{t}^{\mathrm{2}} } {dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(\alpha^{\mathrm{2}} −\mathrm{4}\right){t}^{\mathrm{2}} } \left\{{cos}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right)\:−{isin}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right)\right\}{dt}\:= \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(\alpha^{\mathrm{2}} −\mathrm{4}\right){t}^{\mathrm{2}} } \:{cos}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right)\:−{i}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(\alpha^{\mathrm{2}} −\mathrm{4}\right){t}^{\mathrm{2}} } \:{sin}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right){dt} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(\alpha^{\mathrm{2}} −\mathrm{4}\right){t}^{\mathrm{2}} } {cos}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right){dt}\:=\:\frac{\sqrt{\pi}}{\mathrm{4}+\alpha^{\mathrm{2}} }\:\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(\alpha^{\mathrm{2}} −\mathrm{4}\right){t}^{\mathrm{2}} } {sin}\left(\mathrm{4}\alpha{t}^{\mathrm{2}} \right){dt}\:=\:\frac{\alpha\sqrt{\pi}}{\mathrm{8}\:+\mathrm{2}\alpha^{\mathrm{2}} }\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 17/May/18
let put A(α) = ∫_0 ^∞ e^(−(2+iα)^2 t^2 ) dt changement (2+iα)t =x give A(α) = (1/(2+iα)) ∫_0 ^∞ e^(−x^2 ) dx= ((√π)/(2(2+iα))) = ((√π)/2) ((2−iα)/(4 +α^2 )) = ((√π)/(4+α^2 )) −((α(√π))/(2(4+α^2 )))
$${let}\:{put}\:{A}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left(\mathrm{2}+{i}\alpha\right)^{\mathrm{2}} {t}^{\mathrm{2}} } {dt}\:\:{changement} \\ $$$$\left(\mathrm{2}+{i}\alpha\right){t}\:={x}\:{give} \\ $$$${A}\left(\alpha\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}+{i}\alpha}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−{x}^{\mathrm{2}} } {dx}=\:\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{2}+{i}\alpha\right)} \\ $$$$=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\frac{\mathrm{2}−{i}\alpha}{\mathrm{4}\:+\alpha^{\mathrm{2}} }\:=\:\frac{\sqrt{\pi}}{\mathrm{4}+\alpha^{\mathrm{2}} }\:−\frac{\alpha\sqrt{\pi}}{\mathrm{2}\left(\mathrm{4}+\alpha^{\mathrm{2}} \right)} \\ $$$$ \\ $$
Answered by sma3l2996 last updated on 16/May/18
let u=(2+ia)t⇒dt=(du/(2+ia)) I=∫_0 ^∞ e^(−(2+ia)^2 t^2 ) dt=(1/(2+ia))∫_0 ^∞ e^(−u^2 ) du =((√π)/(2(2+ia)))
$${let}\:{u}=\left(\mathrm{2}+{ia}\right){t}\Rightarrow{dt}=\frac{{du}}{\mathrm{2}+{ia}} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{2}+{ia}\right)^{\mathrm{2}} {t}^{\mathrm{2}} } {dt}=\frac{\mathrm{1}}{\mathrm{2}+{ia}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{2}+{ia}\right)} \\ $$

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