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Study the convergence of $\int_{0}^{+\infty}\ln(1+\frac{1}{t^2})dt$

For $+\infty$ case it's easy we have $\ln(1+\frac{1}{t^2})\sim \frac{1}{t^2}$

For $0$ case I feel it's $o(\frac{1}{\sqrt{t}})$, unfortunately I cannot prove it rigorously.

Thank you in advance,

EDIT: As is pointed on answers, I would like to prove rigorously that $\ln(t^2 + 1) - 2\ln(t)\sim -2ln(t)$. Please do not use l'Hospital rules.

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As Zarrax noted,

$\ln(1+\frac{1}{t^2})=ln(1+t^2)-2ln(t)$

Note that $t^2$ tends to $0$, so $$\frac{\ln(1+\frac{1}{t^2})}{-2\ln(t)}\sim \frac{t^2}{-ln(t^2)}\rightarrow0$$ as $t\rightarrow0$

The result follow.

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Use that $\ln(1 + {1 \over t^2}) = \ln(t^2 + 1) - 2\ln(t)$. The first term is bounded near $t = 0$, so the problem reduces to showing that the integral of $\ln(t)$ converges near $t = 0$.

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  • $\begingroup$ the difference tends to zero, but both go to infinity $\endgroup$ – Zarrax Apr 18 '14 at 21:49
  • $\begingroup$ I still have a problem so..How can I prove that the difference tends to $0$ ? I would like to prove it rigorously..Thanks $\endgroup$ – Free X Apr 18 '14 at 22:00
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Zarrax's method is certainly the one intended, but you could note that an explicit antiderivative is available: $$ F(t) = t \ln\left(1 + \dfrac{1}{t^2}\right) + 2 \arctan(t) $$ and find the limits of $F(t)$ as $t \to 0$ and $t \to \infty$.

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  • $\begingroup$ My problem is to prove rigorously the first limit. $\endgroup$ – Free X Apr 18 '14 at 22:03
  • $\begingroup$ As Zarrax noted, $\ln(1 + 1/t^2) = \ln(t^2+1) - 2 \ln(t)$. You could use l'Hospital on $t \ln t$ as $t \to 0+$. $\endgroup$ – Robert Israel Apr 18 '14 at 23:57

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