Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 83852 by M±th+et£s last updated on 06/Mar/20

$$f\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{e^{-\alpha x}sin\left(x\right)}{x}dx$$

Commented by mathmax by abdo last updated on 06/Mar/20

$$for\alpha ⩾0wehave{f}^{{}^{\prime }}\left(\alpha \right)=-{\int }_0^{\mathrm{\infty }}{e}^{-\alpha x}sinxdx$$$$=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{-\alpha x+ix}dx\right)=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{(-\alpha +i)x}dx\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(-\alpha +i)x}dx={\left[\frac{1}{-\alpha +i}{e}^{(-\alpha +i)x}\right]}_0^{+\mathrm{\infty }}=\frac{1}{\alpha -i}=\frac{\alpha +i}{{\alpha }^2+1}\Rightarrow$$$$f^{\prime }\left(\alpha \right)=-\frac{1}{1+{\alpha }^2}\Rightarrow f\left(\alpha \right)=-arctan\left(\alpha \right)+K$$$$f\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=\frac{\pi }{2}=K\Rightarrow f\left(\alpha \right)=\frac{\pi }{2}-arctan\left(\alpha \right)(\alpha ⩾0)$$

Commented by M±th+et£s last updated on 07/Mar/20

$$thankyousir$$

Commented by abdomathmax last updated on 07/Mar/20

$$youarewelcomesir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com