find-the-value-of-I-0-e-t-sint-t-dt-and-J-0-e-t-cos-t-t-dt-study-first-the-convergence- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 62262 by maxmathsup by imad last updated on 18/Jun/19 findthevalueofI=∫0∞e−tsinttdtandJ=∫0∞e−tcos(t)tdt,studyfirsttheconvergence. Commented by maxmathsup by imad last updated on 19/Jun/19 theconvergenceoIandJisprovedbysirsmailletpassetothevalueswehaveJ−iI=∫0∞e−tt(cost−isint)dt=∫0∞e−te−ittdt=∫0∞e−(i+1)ttdtchangementt=xgiveJ−iI=∫0∞e−(1+i)x2x(2x)dx=2∫0∞e−(1+i)x2dx=∫−∞+∞e−(1+i)x2dx=1+ix=u∫−∞+∞e−u2du1+i=11+i∫−∞+∞e−u2du=π1+i1+i=2eiπ4⇒1+i=42eiπ8⇒J−iI=π(42)e−iπ8=π(42){cos(π8)−isin(π8)}⇒I=π(42)sin(π8)andJ=π(42)cos(π8)⇒I=π(42)2−22andJ=π(42)2+22. Answered by Smail last updated on 18/Jun/19 sint⩽1⇔e−tsint⩽e−te−tsintt⩽e−tt⇔∫0∞e−tsinttdt⩽∫0∞e−ttdtKnowingthat∫0∞e−ttdt=2∫0∞e−x2dx=πwhichmeans∫0∞e−ttdtconverges.ThatmeansI=∫0∞e−tsinttdtconvergesbecauseI⩽πAlsoJ⩽∫0∞e−ttdtmeaningJalsoconvergesJ−iI=∫0∞e−te−ittdt=∫0∞e−t(1+i)tdtLetx=t(1+i)⇒dtt=2dx1+iJ−iI=224eiπ8∫0∞e−x2dx=π24eiπ8=π24(cos(π8)−isin(π8))J=π24cos(π8)=π242+122=J=π(2+1)2I=π(2−1)2 Commented by maxmathsup by imad last updated on 18/Jun/19 thankssir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-127788Next Next post: Question-62263 Leave a Reply Cancel replyYour email address will not be published.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.