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Question Number 181840 by srikanth2684 last updated on 01/Dec/22

$$\underset{-1}{\stackrel{4}{\int }}lnxdx$$

Commented by mr W last updated on 01/Dec/22

$$questioniswrong.$$$$for\ln \left(x\right)tobedefined,x>0!$$

Commented by CElcedricjunior last updated on 01/Dec/22

$${\int }_{-1}^4\mathit{l}\mathit{n}\mathit{x}\mathit{d}\mathit{x}={[\mathit{x}\mathit{l}\mathit{n}\mid \mathit{x}\mid -x]}_{-1}^4=+\mathrm{\infty }$$

Commented by Frix last updated on 01/Dec/22

$$+\mathrm{\infty }\mathrm{is}\mathrm{wrong}!$$$$\int \frac{dx}{x}=\ln \mid x\mid +C$$$$\mathrm{In}\mathrm{this}\mathrm{case}\mathrm{the}\mathrm{absolute}\mathrm{value}\mathrm{makes}\mathrm{sense}$$$$\mathrm{because}\mathrm{obviously}\mathrm{the}\mathrm{area}\mathrm{between}\mathrm{the}$$$$\mathrm{graph}\mathrm{of}y=\frac{1}{x}\mathrm{and}\mathrm{the}x-\mathrm{axis}\mathrm{is}\mathrm{the}\mathrm{same}$$$$\mathrm{in}\mathrm{both}\mathrm{intervals}[a,b]\mathrm{and}[-b,-a]\Rightarrow$$$$\underset{-r}{\stackrel{+r}{\int }}\frac{dx}{x}=0\mathrm{\forall }r\in \mathbb{R}\mathrm{despite}\mathrm{of}\mathrm{the}\mathrm{singularity}.$$$$\int \ln xdx=-x(1-\ln x)+C$$$$\mathrm{In}\mathrm{this}\mathrm{case}y=\ln x\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x⩽0$$$$\mathrm{and}x,y\in \mathbb{R}$$$$\mathrm{If}\mathrm{we}\mathrm{let}x,y\in \mathbb{C}:x<0\wedge r=\mid x\mid \iff x=r{\mathrm{e}}^{\mathrm{i}\pi }\mathrm{and}$$$$\ln x=\ln r+\pi \mathrm{i}\Rightarrow$$$$\underset{-1}{\stackrel{4}{\int }}\ln xdx=-5+8\ln 2+\pi \mathrm{i}$$$$[\mathrm{real}\left(\ln x\right)=\ln \mid x\mid ;\mathrm{imag}\left(\ln x\right)=\frac{\pi (1-\mathrm{sign}x)}{2}]$$

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