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Question Number 46844 by maxmathsup by imad last updated on 01/Nov/18

$$calculate{\int }_0^{\mathrm{\infty }}{e}^{-2t}ln(1+3t)dt$$

Commented by maxmathsup by imad last updated on 04/Nov/18

$$letA={\int }_0^{\mathrm{\infty }}{e}^{-2t}ln(1+3t)dtchangement1+3t=xgivet=\frac{x-1}{3}\Rightarrow$$$$A={\int }_1^{+\mathrm{\infty }}{e}^{-2\frac{x-1}{3}}ln\left(x\right)\frac{dx}{3}=\frac{1}{3}{e}^{\frac{2}{3}}{\int }_1^{+\mathrm{\infty }}{e}^{\frac{-2x}{3}}ln\left(x\right)dxletfindfor\lambda >0$$$$W_{\lambda }={\int }_1^{+\mathrm{\infty }}{e}^{-\lambda x}ln\left(x\right)dxbyparts{W}_{\lambda }={[-\frac{1}{\lambda }{e}^{-\lambda x}ln\left(x\right)]}_1^{+\mathrm{\infty }}+\frac{1}{\lambda }{\int }_1^{+\mathrm{\infty }}{e}^{-\lambda x}\frac{dx}{x}$$$$=\frac{1}{\lambda }{\int }_1^{+\mathrm{\infty }}{e}^{-(\lambda +1)x}dx=\frac{1}{\lambda }{[-\frac{1}{\lambda +1}{e}^{-(\lambda +1)x}]}_1^{+\mathrm{\infty }}=\frac{-1}{\lambda (\lambda +1)}(-e^{-(\lambda +1)})$$$$=\frac{e^{-\lambda -1}}{\lambda (\lambda +1)}\Rightarrow {\int }_1^{+\mathrm{\infty }}{e}^{\frac{-2x}{3}}lnx)dx=\frac{e^{-\frac{2}{3}-1}}{\frac{2}{3}(\frac{2}{3}+1)}=\frac{3}{2}.\frac{3}{5}{e}^{-\frac{5}{3}}$$$$=\frac{9}{10}{e}^{-\frac{5}{3}}\Rightarrow A=\frac{1}{3}{e}^{\frac{2}{3}}\frac{9}{10}{e}^{-\frac{5}{3}}=\frac{3}{10}{e}^{-1}=\frac{3}{10e}.$$

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