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Question Number 32031 by abdo imad last updated on 18/Mar/18

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-ax}ln\left(x\right)dxwitha>0$$$$1)findf\left(a\right)$$$$2)find{\int }_0^{\mathrm{\infty }}{e}^{-ax}\left(xlnx\right)dx$$$$3)calculate{\int }_0^{\mathrm{\infty }}{e}^{-2x}\left(xlnx\right)dx.$$

Commented by abdo imad last updated on 20/Mar/18

$$ch.ax=tgivef\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(\frac{t}{a}\right)\frac{dt}{a}$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}{e}^{-t}(ln\left(t\right)-ln\left(a\right))dt$$$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt-\frac{ln\left(a\right)}{a}{\int }_0^{\mathrm{\infty }}{e}^{-t}dtbutwehaveproved$$$$that{\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt=-\gamma \Rightarrow f\left(a\right)=-\frac{\gamma }{a}-\frac{ln\left(a\right)}{a}$$$$2)wehave{f}^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}x{e}^{-ax}ln\left(x\right)dx\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{-ax}\left(xln\left(x\right)\right)dx=-f^{{}^{\prime }}\left(a\right)fromanotherside$$$$f^{{}^{\prime }}\left(a\right)=\frac{\gamma }{a^2}-\frac{1-ln\left(a\right)}{a^2}=\frac{\gamma +ln\left(a\right)-1}{a^2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{e}^{-ax}\left(xlnx\right)dx=\frac{1-\gamma -ln\left(a\right)}{a^2}$$$$3)fromrel.2)lettakea=2weget$$$${\int }_0^{\mathrm{\infty }}{e}^{-2x}\left(xln\left(x\right)\right)dx=\frac{1-\gamma -ln\left(2\right)}{4}.$$

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