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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.

RFZ
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3 Answers3

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

A.Γ.
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    It's quite nice – RFZ Oct 10 '17 at 07:01
  • Please, take it easy. Prove the convergence of $\int \limits_{0}^{\infty}\left (\dfrac{1}{\sqrt{x^4+4}}-\dfrac{C}{x^2+2}\right)dx$ – Claude Leibovici Oct 10 '17 at 07:45
  • @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? – A.Γ. Oct 10 '17 at 07:51
  • 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 ? – Claude Leibovici Oct 10 '17 at 08:00
  • @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. – A.Γ. Oct 10 '17 at 08:03
  • I am afraid that this was not the question. – Claude Leibovici Oct 10 '17 at 08:04
  • @ClaudeLeibovici Sorry, what was the question? Antiderivatives exist, the problem is to find them in elementary terms. Here it is possible. – A.Γ. Oct 10 '17 at 08:07
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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.

Claude Leibovici
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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 $\lim_{x\to\infty}(1-C\sqrt{1+\frac{4}{x^2}})=1-C$ $\Rightarrow C=1$

samer abdullah
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