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Question Number 82446 by M±th+et£s last updated on 21/Feb/20

$$showthat$$$${\int }_0^{\mathrm{\infty }}{x}^{-log\left(x\right)}xlog\left(x\right)dx=e\sqrt{\pi }$$

Commented by abdomathmax last updated on 21/Feb/20

$$changementlogx=tgivex=e^t\Rightarrow$$$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\left(e^t\right)}^{-t}{e}^{t}t{e}^{t}dt$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}t{e}^{-t^2+2t}dt={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}t{e}^{-(t^2-2t+1-1)}dt$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}t{e}^{-{(t-1)}^2+1}dt{=}_{t-1=u}e{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(u+1)e^{-u^2}du$$$$but{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(u+1)e^{-u^2}du$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}u{e}^{-u^2}du+{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}du$$$$={[-\frac{1}{2}{e}^{-u^2}]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\sqrt{\pi }=0+\sqrt{\pi }\Rightarrow$$$$I=e\sqrt{\pi }$$

Commented by M±th+et£s last updated on 21/Feb/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 21/Feb/20

$$youarewelcome.$$

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