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Question:

$\int ^1_0 \frac {\ln x}{1-x^2}dx$ - converges or diverges?

What we did:

We tried to compare with $-\frac 1x$ and $-\frac 1{x-1}$ but ended up finding that these convergence tests fail. Our book says this integral diverges, but Wolfram on the other hand says it converges. How come?

jimjim
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jreing
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  • I think the book is wrong. First compare the integrand to $\log x$ near $x=0$; then see what happens to the integrand near $x=1$. – Gerry Myerson Oct 26 '13 at 11:46
  • You might be interested in this question it doesn't tackle convergence as such but there are some nice ways of evaluating the integral itself http://math.stackexchange.com/questions/537903/proving-int-01-frac-lnxx2-1dx-frac-pi28 – Graham Hesketh Oct 26 '13 at 18:24

2 Answers2

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There are two potential sources of divergence if the integral were to diverge (it doesn't): at $x=0$ and $x=1$. At $x=0$, the integral behaves as $\ln{x}$, which has antiderivative $x \ln{x}-x$. You may show that the limit of this expression as $x \to 0$ is $0$ using e.g.,L'Hopital. So the integrand is integrable at $x=0$.

At $x=1$, you may show that

$$\lim_{x\to 1} \frac{\ln{x}}{1-x^2} = -\frac12$$

so that the integrand is integrable here as well. Thus, the integral converges.

Ron Gordon
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  • Hi. Thank you very much for the answer, It really helped. One last question about your way- when x=1 did you use the limit comparison test with g(x)=1 (which converges) and that's how you got your limit that equaled -0.5? – jreing Oct 29 '13 at 06:09
  • @user1685224: I used the fact that $\ln{x} \sim x-1$ as $x \to 1$. – Ron Gordon Oct 29 '13 at 07:47
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It converges and its value is $-\pi^2/8$.

Riemann1337
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  • $-\frac{\pi^2}8$ – Lucian Oct 26 '13 at 11:51
  • I am not sure how this could be of any interest. If you change variable x = Exp[y], your integrand becomes "-y Csch[y] dy", to be integrated between minus infinity and zero. The antiderivative is not very complex. As mentioned by Lucian, the value of the integral is (- Pi^2 / 8) – Claude Leibovici Oct 26 '13 at 12:38
  • To be a complete answer you need to say why... – user1729 Oct 26 '13 at 12:56