$\int_0^1 \ln(1+\frac{a^2}{x^2})dx $
How should i prove that this integral converges?
This is what i did:
$u=\frac{1}{x}$ and $dx=-\frac{1}{u^2}du$
$\int_0^1 \ln(1+\frac{a^2}{x^2})dx=\int_{\infty}^{1} \ln(1+\frac{a^2}{\frac{1}{u^2}})(-\frac{1}{u^2})du= \int_{1}^{\infty}\frac{\ln(1+a^2u^2)}{u^2} $
Im not sure if this helps at all, i know that
$\lim_{u\to\infty}\frac{\ln(1+a^2u^2)}{u^2}$ equals $\lim_{u\to\infty}\frac{2\ln(au)}{u^2}$
but i dont know on what to compare that...
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Valentin
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Hint: Argue that $\frac{\ln(1 + a^2 u^2)}{u^2} \leq \frac{C}{u^{\alpha}}$ for appropriate constants $C$ and $\alpha$ such that $\int_1^{\infty} \frac{C}{u^{\alpha}} , du$ converges. – Michael Joyce Nov 26 '15 at 15:38
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I dont understand. How could i do that? If $\ln (1+a^2u^2) \to \infty$ – Valentin Nov 26 '15 at 21:00
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But, e.g., $\frac{\ln(1+a^2u^2)}{\sqrt{u}} \rightarrow 0$ as $u \rightarrow \infty$. – Michael Joyce Nov 26 '15 at 22:06
2 Answers
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I do not know if I am off-topic. If this is the case, please forgive me.
May be, we could compute the antiderivative by parts $$u=\log(1+\frac {a^2}{x^2}) \implies du=-\frac{2 a^2}{a^2 x+x^3}dx$$ $$dv=dx \implies v=x$$ $$\int \log(1+\frac{a^2}{x^2})\,dx=x \log \left(1+\frac{a^2}{x^2}\right)+\int\frac{2 a^2}{a^2+x^2}\,dx$$ The second integral is quite simple $(x=a y)$ and you then have $$\int \log(1+\frac{a^2}{x^2})\,dx=x \log \left(1+\frac{a^2}{x^2}\right)+2 a \tan ^{-1}\left(\frac{x}{a}\right)$$
Claude Leibovici
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Claude, how are you? This seems like the best way to show convergence to me! +1 – Mark Viola Nov 26 '15 at 15:42
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I know thats the easyest way, but i should do it without calculating the integral. I mean by limits or comparison thanks – Valentin Nov 26 '15 at 15:46
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@Valentin. The problem is around $x=0$. Use Taylor to get $$\log(1+\frac{a^2}{x^2})=\left(\log \left(a^2\right)-2 \log (x)\right)+\frac{x^2}{a^2}+O\left(x^4\right)$$ and integrate to get, close to $x=0$, $$x \left(\log \left(a^2\right)-2 \log (x)+2\right)+\frac{x^3}{3 a^2}+O\left(x^5\right)$$ which behaves pretty well at $x=0$. – Claude Leibovici Nov 26 '15 at 15:53
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If $a=0$ your claim is trivial. Otherwise, if wlog $a> 0$, setting $y=a/x$, you obtain $$ \int_0^1 \ln\left(1+\frac{a^2}{x^2}\right)dx=a\int_a^\infty \frac{ln(1+y^2)}{y^2}dy. $$ It follows that for every $\varepsilon>0$ it holds $$ \int_0^1 \ln\left(1+\frac{a^2}{x^2}\right)dx =\mathcal{O}\left( \int_a^\infty \frac{y^\varepsilon}{y^2}dy\right), $$ which is finite, whenever $\varepsilon \in ]0,1[$.
Paolo Leonetti
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Is that becouse for $y>y_0 $ you know that $ \ln y^2< y^{\varepsilon}$?? – Valentin Nov 26 '15 at 16:17
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yes, to be precise for $y$ large you have $\ln (1+y^2) \le 3\ln(y) \le y^\varepsilon$ – Paolo Leonetti Nov 26 '15 at 16:19
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