Question Number 28756 by abdo imad last updated on 29/Jan/18
$${find}\:{in}\:{terms}\:{of}\:\lambda\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\lambda{t}} \:\frac{{sint}}{\:\sqrt{{t}}}\:{dt}\:\:{with}\:\:\lambda>\mathrm{0} \\ $$
Commented by abdo imad last updated on 30/Jan/18
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{t}} \:\:\frac{{sint}}{\:\sqrt{{t}}}{dt}\:\:\:{with}\:\:\lambda>\mathrm{0}\:{the}\:{ch}.\sqrt{{t}}\:={x}\:{give} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:\:} \frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}}\:\mathrm{2}{xdx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{e}^{−\lambda{x}^{\mathrm{2}} } \:{sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:{e}^{−\lambda{x}^{\mathrm{2}} } \:{sin}\left({x}^{\mathrm{2}} \right){dx}=−\:{Im}\:\left(\int_{−\infty} ^{+\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:−{ix}^{\mathrm{2}} } {dx}\right)\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{e}^{−\lambda{x}^{\mathrm{2}} \:−{ix}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\lambda+{i}\right){x}^{\mathrm{2}} } {dx}=\:\int_{−\infty} ^{+\infty} \:{e}^{−\left(\sqrt{\lambda+{i}}{x}\right)^{\mathrm{2}} } \:{dx}\:\:\: \\ $$$$=\:\int_{−\infty\:} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{du}}{\:\sqrt{\lambda+{i}}}\:\:\:\:\:\:\:\:\left({ch}.\sqrt{\lambda+{i}}\:{x}={u}\right) \\ $$$$=\left(\lambda\:+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} .\sqrt{\pi}\:\:\:\:\:\:\:\:\:{we}\:{have} \\ $$$$\lambda+{i}=\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\left(\:\:\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} \:}}\:\:+\frac{{i}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\right)\:={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$$${and}\:{cos}\theta=\:\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} \:}}\:\:\:\:{and}\:{sin}\theta=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:}\:\:\Rightarrow{tan}\theta=\frac{\mathrm{1}}{\lambda} \\ $$$$\Rightarrow\theta=\:{arctan}\left(\frac{\mathrm{1}}{\lambda}\right)\:{so}\:\lambda+{i}=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{{iarctan}\left(\frac{\mathrm{1}}{\lambda}\right)} \\ $$$$\lambda+{i}=\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}\:−{arctan}\lambda\right)} \\ $$$$\Rightarrow\:\left(\lambda+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\left(\:\frac{−\pi}{\mathrm{4}}\:+\frac{{arctan}\lambda}{\mathrm{2}}\right)} \\ $$$$=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\left({cos}\left(\:\frac{−\pi}{\mathrm{4}}\:+\frac{{arctan}\lambda}{\mathrm{2}}\right)\:+{isin}\left(\frac{−\pi}{\mathrm{4}}+\frac{{arctan}\lambda}{\mathrm{2}}\right)\right) \\ $$$${I}=−{Im}\left(\sqrt{\pi}\left(\lambda+{i}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)=−\sqrt{\pi}{sin}\left(\frac{−\pi}{\mathrm{4}}\:+\frac{{artan}\lambda}{\mathrm{2}}\right) \\ $$$$=\sqrt{\pi}\:{sin}\left(\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\lambda}{\mathrm{2}}\right)\:. \\ $$