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$$\displaystyle \int_{0.5}^1 \frac{\ln(1-t^2)}{t^2} \mathrm{d}t = \lim_{h\to 1^-} \int_{0.5}^h \frac{\ln(1-t^2)}{t^2} \mathrm{d}t =$$

$$ \lim_{h\to 1^-} \left[\ln(1-t)-\ln(1+t)-\frac{\ln(1-t^2)}{t} \right]_{0.5}^{h} = -\infty - \ln 2 +\infty \cdots$$

The $$-\infty +\infty $$ in the solution has me stumped. How do I get rid of this indeterminacy? (Or did I make a mistake somewhere else? I have checked it twice.)

Even when I rewrite the solution the indeterminacy is still there: $$= \lim_{h\to 1^-} \left[\ln(1-t)-\ln(1+t)-\frac{\ln(1-t)}{t} -\frac{\ln(1+t)}{t} \right]_{0.5}^{h}$$

Thanks for your help.

Michael Hardy
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TylerC
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  • You've already asked a related question - http://math.stackexchange.com/questions/740156/limits-what-if-it-evaluates-to-infty-infty –  Apr 05 '14 at 02:57

2 Answers2

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Hint

You also can rewrite the antiderivative $$f(t)=\ln(1-t)-\ln(1+t)-\frac{\ln(1-t)}{t} -\frac{\ln(1+t)}{t} $$ as $$f(t)=-\frac{(1+t) \log (1+t)}{t}-\frac{(1-t) \log (1-t)}{t}$$ and remember that, when $x$ goes to $0$, $x \log(x)$ goes to $0$ too.

So,$f(1)=-2 \log(2)$ and $f(\frac {1}{2})=-3 \log(3)+4 \log(2)$ and the value of the integral is then $3 \log(3)-4 \log(2)$

Claude Leibovici
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Hint: $$\lim_{h\to 1^-} \Big[\ln(1-t)-\ln(1+t)-\frac{\ln(1-t^2)}{t} \Big]_{0.5}^{h}=\lim_{h\to 1^-} \Big[\ln\left(\dfrac{1-t}{1+t}\right)-{\ln((1-t^2)^{1/t})} \Big]_{0.5}^{h}$$ Use the limit laws, and then evaluate each limit separately.

  • How does this solve the problem? The limit of both components is $\infty$. This tells you nothing about the limit of the difference. – Christopher K Apr 05 '14 at 04:46