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Question Number 61809 by aliesam last updated on 09/Jun/19

Commented by maxmathsup by imad last updated on 10/Jun/19

$$letA={\int }_0^1\frac{x^n}{\sqrt{ln\left(x\right)}}dx\Rightarrow A={\int }_0^1\frac{x^n}{\sqrt{-(-lnx)}}dx=\frac{1}{\sqrt{-1}}{\int }_0^1\frac{x^n}{\sqrt{-ln\left(x\right)}}dx$$$$changement\sqrt{-ln\left(x\right)}=tgive-ln\left(x\right)=t^2\Rightarrow ln\left(x\right)=-t^2\Rightarrow x=e^{-t^2}\Rightarrow$$$$A=\frac{1}{i}{\int }_{+\mathrm{\infty }}^0\frac{e^{-{nt}^2}}{t}(-2t)e^{-t^2}dt=\frac{2}{i}{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t^2}dt=-2i{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t^2}dt$$$${=}_{\sqrt{n+1}t=u}-2i{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{n+1}}=\frac{-2i}{\sqrt{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=\frac{-2i}{\sqrt{n+1}}\frac{\sqrt{\pi }}{2}\Rightarrow$$$$A=\frac{-i\sqrt{\pi }}{\sqrt{n+1}}.$$

Answered by perlman last updated on 09/Jun/19

$$\sqrt{ln\left(x\right)}=i\sqrt{-ln\left(x\right)}\mathrm{\forall }x\in ]0;1]$$$$withe\sqrt{-ln\left(x\right)}=y$$$$x=e^{-y^2}$$$$={\int }_{+\mathrm{\infty }}^0\frac{e^{-{ny}^2}}{iy}(-2y)e^{-y^2}dy$$$$=i{\int }_0^{+\mathrm{\infty }}{e}^{-(n+1)y^2}dy$$$$letz=\sqrt{(n+1)}y$$$$dz=\sqrt{n+1}dy$$$$=i{\int }_0^{+\mathrm{\infty }}\sqrt{n+1}{e}^{-{\mathit{z}}^2}\mathit{d}\mathit{z}=\mathit{i}\sqrt{\mathit{n}+1}{\int }_0^{+\mathrm{\infty }}{e}^{-z^2}dz=i\sqrt{n+1}\frac{\sqrt{\pi }}{2}$$$$$$

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