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$\int_0^\infty\ln(\frac{1+x}{1-x}) \frac{dx}{x}$ can this integral be done by power series If yes I hope a full solution I dont need to use by parts

Mohammad Hussein
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  • The integral doesn't converge, obviously, maybe the upper limit of integration was meant to be $1$. If you want to have your homework done for you, you might at least copy it accurately. –  Jun 25 '17 at 05:38
  • I cant write well in math jax I meant the abdolute value – Mohammad Hussein Jun 25 '17 at 06:07

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For $x>1$, $(1+x)/(1-x)<0$ so its logarithm would have an imaginary part. If that is $\pi i$, the integral has imaginary part $\pi i\int_1^\infty dx/x$ which diverges.

Maybe you wanted to evaluate $$\int_1^\infty\ln\left|\frac{1+x}{1-x}\right|\frac{dx}x$$ instead? You can break this into integrals over $(0,1)$ and $(1,\infty)$. In the latter, use the substitution $y=1/x$ to get $$\int_1^\infty\ln\left|\frac{1+x}{1-x}\right|\frac{dx}x =\int_0^1\ln\left|\frac{y+1}{y-1}\right|\frac{dy}y$$ which is the same as the integral over $(0,1)$. The integral over $(0,1)$ is $$\int_0^1\left(2\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}\right)\frac{dx}x$$ etc.

Angina Seng
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