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Question Number 28756 by abdo imad last updated on 29/Jan/18

$$findintermsof\lambda {\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}\frac{sint}{\sqrt{t}}dtwith\lambda >0$$

Commented byabdo imad last updated on 30/Jan/18

$$letputI={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}\frac{sint}{\sqrt{t}}dtwith\lambda >0thech.\sqrt{t}=xgive$$ $$I={\int }_0^{\mathrm{\infty }}{e}^{-\lambda x^2}\frac{sin\left(x^2\right)}{x}2xdx=2{\int }_0^{\mathrm{\infty }}{e}^{-\lambda x^2}sin\left(x^2\right)dx$$ $$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\lambda x^2}sin\left(x^2\right)dx=-Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\lambda x^2-{ix}^2}dx\right)but$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-\lambda x^2-{ix}^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-(\lambda +i)x^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{\left(\sqrt{\lambda +i}x\right)}^2}dx$$ $$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{\lambda +i}}(ch.\sqrt{\lambda +i}x=u)$$ $$={(\lambda +i)}^{-\frac{1}{2}}.\sqrt{\pi }wehave$$ $$\lambda +i=\sqrt{1+{\lambda }^2}(\frac{\lambda }{\sqrt{1+{\lambda }^2}}+\frac{i}{\sqrt{1+{\lambda }^2}})=r{e}^{i\theta }\Rightarrow r=\sqrt{1+{\lambda }^2}$$ $$andcos\theta =\frac{\lambda }{\sqrt{1+{\lambda }^2}}andsin\theta =\frac{1}{\sqrt{1+{\lambda }^2}}\Rightarrow tan\theta =\frac{1}{\lambda }$$ $$\Rightarrow \theta =arctan\left(\frac{1}{\lambda }\right)so\lambda +i={(1+{\lambda }^2)}^{\frac{1}{2}}{e}^{iarctan\left(\frac{1}{\lambda }\right)}$$ $$\lambda +i={(1+{\lambda }^2)}^{\frac{1}{2}}{e}^{i(\frac{\pi }{2}-arctan\lambda )}$$ $$\Rightarrow {(\lambda +i)}^{-\frac{1}{2}}={(1+{\lambda }^2)}^{-\frac{1}{4}}{e}^{i(\frac{-\pi }{4}+\frac{arctan\lambda }{2})}$$ $$={(1+{\lambda }^2)}^{-\frac{1}{4}}(cos(\frac{-\pi }{4}+\frac{arctan\lambda }{2})+isin(\frac{-\pi }{4}+\frac{arctan\lambda }{2}))$$ $$I=-Im\left(\sqrt{\pi }{(\lambda +i)}^{-\frac{1}{2}}\right)=-\sqrt{\pi }sin(\frac{-\pi }{4}+\frac{artan\lambda }{2})$$ $$=\sqrt{\pi }sin(\frac{\pi }{4}-\frac{arctan\lambda }{2}).$$

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