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Find the value of the constant $C$ for which the integral $$\int \limits_{0}^{\infty}\left (\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\right)dx$$ converges. Evaluate the integral for this value of $C$.

I have some difficulties with above problem. I know some methods such as $x+2 \sim x$ and $\sqrt{x^2+4}\sim x$ for $x \to \infty$. But I would like to see the rigorous proof.

Can anyone please show it?

Would very thankful for that.

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Hint: $$ \int_0^a\Big(\frac{1}{\sqrt{x^2+4}}-\frac{C}{x+2}\Big)\,dx=\left[\ln(x+\sqrt{x^2+4})-C\ln(x+2)\right]_0^a=\ln\left(\frac{a+\sqrt{a^2+4}}{(a+2)^C}\right)+\text{const} $$

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    $\begingroup$ It's quite nice $\endgroup$ – ZFR Oct 10 '17 at 7:01
  • $\begingroup$ Please, take it easy. Prove the convergence of $\int \limits_{0}^{\infty}\left (\dfrac{1}{\sqrt{x^4+4}}-\dfrac{C}{x^2+2}\right)dx$ $\endgroup$ – Claude Leibovici Oct 10 '17 at 7:45
  • $\begingroup$ @ClaudeLeibovici This is what one can do when calculate the limit $a\to+\infty$. It proves convergence by definition. Don't you agree that it is easier than to study asymptotics? $\endgroup$ – A.Γ. Oct 10 '17 at 7:51
  • $\begingroup$ I totally agree with you using $\int_0^a f(x)\,dx$ and then to look at the limit when $a\to \infty$. But, you need to find the antiderivative first. What if it does not exist ? $\endgroup$ – Claude Leibovici Oct 10 '17 at 8:00
  • $\begingroup$ @ClaudeLeibovici Then it will be another exercise :) Here you have to evaluate the integral anyway as a part of the question, so calculating the antiderivatives are necessary sooner or later. $\endgroup$ – A.Γ. Oct 10 '17 at 8:03
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Look what happens close to the bounds using Taylor expansions.

Close to $x=0$ $$\frac{1}{\sqrt{x^2+4}}-\frac{c}{x+2}=\left(\frac{1}{2}-\frac{c}{2}\right)+\frac{c x}{4}+O\left(x^2\right)$$

For large values of $x$ $$\frac{1}{\sqrt{x^2+4}}-\frac{c}{x+2}=\frac{1-c}{x}+O\left(\frac{1}{x^2}\right)$$

I am sure that you can conclude from here.

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  • $\begingroup$ How did you get the second expansion? And why do you consider two cases? $\endgroup$ – ZFR Oct 10 '17 at 6:57
  • $\begingroup$ For the second one, factor $\frac 1x$ and expand. Why . Just because we need to check at both bounds (even if in this case, it is obvious that there is no problem around $x=0$. Is this clear ? $\endgroup$ – Claude Leibovici Oct 10 '17 at 7:20
  • $\begingroup$ Yes it is clear. But I dont know what to do next with your expansions? $\endgroup$ – ZFR Oct 10 '17 at 7:33
  • $\begingroup$ @RFZ. Try to integrate the expansions close to the bounds. $\endgroup$ – Claude Leibovici Oct 10 '17 at 7:43
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First we have,

$\int_0^\infty \frac{1}{\sqrt{x^+4}}-\frac{C}{x+2}$ $=\int_0^\infty \frac{x+2-C\sqrt{x^2+4}}{(x+2)(\sqrt{x^2+4}}$ $=\int_0^\infty \frac{x+2-Cx\sqrt{1+\frac{4}{x^2}}}{(x^2+2x)(\sqrt{1+\frac{4}{x^2}}})$ $=\int_0^\infty \frac{x(1-C\sqrt{1+\frac{4}{x^2}})+1}{x^2\sqrt{1+\frac{4}{x^2}}+x\sqrt{1+\frac{4}{x^2}}}$ For this to converge we must have the coefficient of the x in the numerator going to 0 as $x\to\infty$(otherwise the integral will diverge by limit comparison with $\int\frac{1}{x}$).SO $\lim_{x\to\infty}(1-C\sqrt{1+\frac{4}{x^2}})=1-C$ $\Rightarrow C=1$

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