Menu Close

let-f-a-0-e-ax-ln-x-dx-with-a-gt-0-1-find-f-a-2-find-0-e-ax-xlnx-dx-3-calculate-0-e-2x-xlnx-dx-

Tinku Tara 0 Comments
Pin




Question Number 32031 by abdo imad last updated on 18/Mar/18
let f(a) = ∫_0 ^∞ e^(−ax) ln(x)dx with a>0 1) find f(a) 2) find ∫_0 ^∞ e^(−ax) (xlnx)dx 3) calculate ∫_0 ^∞ e^(−2x) (xlnx)dx .
letf(a)=∫0∞e−axln(x)dxwitha>01)findf(a)2)find∫0∞e−ax(xlnx)dx3)calculate∫0∞e−2x(xlnx)dx.
Commented by abdo imad last updated on 20/Mar/18
ch. ax =t give f(a) =∫_0 ^∞ e^(−t) ln ((t/a))(dt/a) =(1/a) ∫_0 ^∞ e^(−t) (ln(t) −ln(a))dt = (1/a)∫_0 ^∞ e^(−t) ln(t)dt −((ln(a))/a) ∫_0 ^∞ e^(−t) dt but we have proved that ∫_0 ^∞ e^(−t) ln(t)dt =−γ ⇒ f(a) =−(γ/a) −((ln(a))/a) 2) we have f^′ (a) = −∫_0 ^∞ x e^(−ax) ln(x)dx ⇒ ∫_0 ^∞ e^(−ax) (xln(x))dx = −f^′ (a) from another side f^′ (a) = (γ/a^2 ) − ((1 −ln(a))/a^2 ) = ((γ +ln(a) −1)/a^2 ) ⇒ ∫_0 ^∞ e^(−ax) (xlnx)dx= ((1−γ −ln(a))/a^2 ) 3) from rel. 2) let take a=2 we get ∫_0 ^∞ e^(−2x) (xln(x))dx = ((1−γ −ln(2))/4) .
ch.ax=tgivef(a)=∫0∞e−tln(ta)dta=1a∫0∞e−t(ln(t)−ln(a))dt=1a∫0∞e−tln(t)dt−ln(a)a∫0∞e−tdtbutwehaveprovedthat∫0∞e−tln(t)dt=−γ⇒f(a)=−γa−ln(a)a2)wehavef′(a)=−∫0∞xe−axln(x)dx⇒∫0∞e−ax(xln(x))dx=−f′(a)fromanothersidef′(a)=γa2−1−ln(a)a2=γ+ln(a)−1a2⇒∫0∞e−ax(xlnx)dx=1−γ−ln(a)a23)fromrel.2)lettakea=2weget∫0∞e−2x(xln(x))dx=1−γ−ln(2)4.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *