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Question Number 72025 by mathmax by abdo last updated on 23/Oct/19

$$calculate{\int }_0^{\mathrm{\infty }}{e}^{-\alpha x}ln\left(x\right)dxwith\alpha >0$$

Commented by mathmax by abdo last updated on 24/Oct/19

$$changement\alpha x=tgive$$$${\int }_0^{\mathrm{\infty }}{e}^{-\alpha x}ln\left(x\right)dx={\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(\frac{t}{\alpha }\right)\frac{dt}{\alpha }=\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}{e}^{-t}\{ln\left(t\right)-ln\alpha \}dt$$$$=\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)-\frac{ln\left(\alpha \right)}{\alpha }$$$$=\frac{1}{\alpha }(-\gamma )-\frac{ln\left(\alpha \right)}{\alpha }=-\frac{1}{\alpha }(\gamma +ln\left(\alpha \right))$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}ln\left(t\right)dt=-\gamma andthisvalueisproved.$$

Answered by mind is power last updated on 23/Oct/19

$$\mathrm{u}=\alpha \mathrm{x}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\ln \left(\frac{\mathrm{u}}{\alpha }\right).\frac{\mathrm{du}}{\alpha }$$$$={\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\ln \left(\mathrm{u}\right)\frac{\mathrm{du}}{\alpha }-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}.\frac{\ln \left(\alpha \right)}{\alpha }\mathrm{du}$$$$=\frac{1}{\alpha }{\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\ln \left(\mathrm{u}\right)\mathrm{du}+\frac{\ln \left(\alpha \right)}{\alpha }{\left[{\mathrm{e}}^{-\mathrm{u}}\right]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{\alpha }.\gamma -\frac{\ln \left(\alpha \right)}{\alpha }=\frac{1}{\alpha }(\gamma -\ln \left(\alpha \right))$$

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