I would like to compute numerically the definite integral below using the Simpson rule:
$$\int\limits_0^1 \ln (x) \ln (1-x)\ dx$$
But I have difficulty trying to remove the singularity in this integral.
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1 Answer
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If you have a singularity with natural log, then your solution is imaginary. The indefinite integral of $\ln(x)\ln(x-1)$ is $$x((\ln(x-1)-1)\ln(x)-\ln(x-1)+2)+\ln(x-1)+ Li_2 (1-x)+C$$Using integral-calculator.com . Li_2 is $$\displaystyle\sum^{\infty}_{k=1} \frac{x^k}{k^2}$$ Now note that the natural log of -1 is $\pi i$, so our final answer is $$\pi i -2+\frac{\pi^2}{6}$$ Hope this helps :)
