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I would like to evaluate the integral $\int_0^1\ln(\frac{1+x}{1-x})dx$ by expanding the integrands into power series. Asking wolframalpha the answer should be $2\ln(2)$ but I don't get there. What I tried so far:

I know that $\ln(x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}(x-1)^k$ if $|x-1|\leq 1$ and $x\neq 0$. We also have $\ln(\frac{1+x}{1-x})=\ln(1+x)-\ln(1-x)$. Using the formula for the power series, I get

$\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}x^k$ on $[0,1]$ and $\ln(1-x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}(-x)^k=-\sum_{k=1}^\infty \frac{1}{k}x^k$ on $[0,1)$. Can I just substract this know? Is it right that then

$\ln(\frac{1+x}{1-x})=\ln(1+x)-\ln(1-x)=2\sum_{k=0}^\infty \frac{x^{2k+1}}{2k+1}$? I am not sure if this is right and how can I proceed from there?

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  • $\begingroup$ Integrate both sides. $\endgroup$
    – Symposium
    Aug 21, 2018 at 12:43
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    $\begingroup$ Yes, that is correct. Those series converge absolutely for $|x|<1$, so you can combine the series, then integrate term by term. $\endgroup$
    – Clayton
    Aug 21, 2018 at 12:44
  • $\begingroup$ @Clayton Thanks! At least I was right until there… Now, integrating term by term, I get $2\sum_{k=0}^\infty \frac{x^{2k+2}}{(2k+2)(2k+1)}$ right? Now, plugging in 1 and 0 (which gives zero). So I get $\int_0^1 \ln(\frac{1+x}{1-x})=2\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+1)}$? How do I see that the sum is $\ln(2)$? I know that $\ln(2)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}$. $\endgroup$
    – Algebra
    Aug 21, 2018 at 12:57

1 Answer 1

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So far your ideas are correct. You can write $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x)$ and then write each logarithm in its respective power series. Because it is a power series and $|x|<1$ (where the radius of convergence is $1$ in this case), you can combine the power series (rearrangement doesn't matter for absolutely convergent series). Now when you integrate, you'll end up with $$2\sum_{k=0}^\infty\frac{1}{(2k+1)(2k+2)}=2\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)=2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}=2\ln(2).$$ The first identity follows from partial fractions and the last identity follows from the convergence of the power series for $\ln(1+x)$ when $x=1$.

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