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Question Number 212368 by Spillover last updated on 12/Oct/24

Answered by MathematicalUser2357 last updated on 13/Oct/24

$$\mathrm{\infty }$$

Commented by Ghisom last updated on 14/Oct/24

$$\mathrm{no}$$

Commented by MathematicalUser2357 last updated on 04/Nov/24

$$\mathrm{Then}\mathrm{calculate}{\int }_{-1}^1\ln xdx???$$

Commented by Ghisom last updated on 04/Nov/24

$$\underset{-1}{\stackrel{1}{\int }}\ln xdx={[x\ln x-x]}_{-1}^1=-2+\pi \mathrm{i}$$$$\mathrm{but}\mathrm{here}\mathrm{we}\mathrm{have}$$$$\underset{-1}{\stackrel{1}{\int }}\ln x^2\ln (1-x^2)dx=$$$$=16-{\pi }^2-8\ln 2\approx .585218154431$$

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