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Question Number 64200 by aliesam last updated on 15/Jul/19

Commented by mathmax by abdo last updated on 15/Jul/19

$$letI={\int }_0^1\frac{x-1}{lnx}dxchangementlnx=-tgivex=e^{-t}$$$$I=-{\int }_0^{\mathrm{\infty }}\frac{e^{-t}-1}{-t}(-e^{-t})dt$$$$=-{\int }_0^{\mathrm{\infty }}\frac{e^{-2t}-e^{-t}}{t}dtletintroducetheparametricfunction$$$$f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-2t}-e^{-t}}{t}{e}^{-xt}dtwithx>0wehave$$$$f^{{}^{\prime }}\left(x\right)=-{\int }_0^{\mathrm{\infty }}(e^{-2t}-e^{-t})e^{-xt}dt=-{\int }_0^{\mathrm{\infty }}\{e^{-(x+2)t}-e^{-(x+1)t}\}dt$$$$={\int }_0^{\mathrm{\infty }}(e^{-(x+1)t}-e^{-(x+2)t})dt={[-\frac{1}{x+1}{e}^{-(x+1)t}+\frac{1}{x+2}{\mathrm{e}}^{-(\mathrm{x}+2)\mathrm{t}}]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{x+1}-\frac{1}{x+2}\Rightarrow f\left(x\right)=ln(x+1)-ln(x+2)+c=ln\left(\frac{x+1}{x+2}\right)+c$$$$\mathrm{\exists }m>0/\mid f\left(x\right)\mid ⩽m{\int }_0^{\mathrm{\infty }}{e}^{-xt}dt=\frac{m}{x}\to 0(x\to +\mathrm{\infty })$$$$c={lim}_{x\to +\mathrm{\infty }}(f\left(x\right)-ln\left(\frac{x+1}{x+2}\right))=0f\left(x\right)=ln\left(\frac{x+1}{x+2}\right)$$$$I={lim}_{x\to 0}-f\left(x\right)=ln\left(2\right).$$

Answered by Hope last updated on 15/Jul/19

$$tricky$$$${\int }_0^1\frac{x^a-1}{lnx}=f\left(a\right)$$$$\frac{df}{da}={\int }_0^1\frac{\partial }{\partial a}\frac{x^a-1}{lnx}dx$$$$={\int }_0^1\frac{x^{a}lnx}{lnx}dx$$$${\int }_0^1{x}^{a}dx$$$$=\mid \frac{x^{a+1}}{a+1}{\mid }_0^1=\frac{1}{a+1}$$$$\frac{df}{da}=\frac{1}{a+1}$$$$df=\frac{da}{a+1}$$$$f\left(a\right)=ln(a+1)+c$$$$whena=0f\left(0\right)=0soc=0$$$$f\left(a\right)=ln(a+1)$$$$soansisf\left(1\right)$$$$=ln2$$$$$$$$$$

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