f-0-e-x-sin-x-x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 83852 by M±th+et£s last updated on 06/Mar/20 f(α)=∫0∞e−αxsin(x)xdx Commented by mathmax by abdo last updated on 06/Mar/20 forα⩾0wehavef′(α)=−∫0∞e−αxsinxdx=−Im(∫0∞e−αx+ixdx)=−Im(∫0∞e(−α+i)xdx)but∫0∞e(−α+i)xdx=[1−α+ie(−α+i)x]0+∞=1α−i=α+iα2+1⇒f′(α)=−11+α2⇒f(α)=−arctan(α)+Kf(0)=∫0∞sinxxdx=π2=K⇒f(α)=π2−arctan(α)(α⩾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 FacebookTweetPin Post navigation Previous Previous post: Question-149385Next Next post: In-a-triangle-ABC-with-fixed-base-BC-the-vertex-A-moves-such-that-cos-B-cos-C-4-sin-2-A-2-If-a-b-and-c-denote-the-lengths-of-the-sides-of-the-triangle-opposite-to-the-angles-A-B-and-C-resp Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![for α≥0 we have f^′ (α) =−∫_0 ^∞ e^(−αx) sinx dx =−Im(∫_0 ^∞ e^(−αx+ix) dx) =−Im(∫_0 ^∞ e^((−α+i)x) dx) but ∫_0 ^∞ e^((−α+i)x) dx =[(1/(−α +i)) e^((−α+i)x) ]_0 ^(+∞) =(1/(α−i)) =((α+i)/(α^2 +1)) ⇒ f′(α) =−(1/(1+α^2 )) ⇒f(α)=−arctan(α)+K f(0) =∫_0 ^∞ ((sinx)/x)dx =(π/2) =K ⇒f(α)=(π/2) −arctan(α) (α≥0)](https://www.tinkutara.com/question/Q83856.png)
