Determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent So, I need to determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent, for any $ C\in \mathbb{R} $.
I found that for C=1 the integral convergent. I think its easy to show that for $ C\leq 0 $ the integral will diverge, but I dont know how to determine for the rest of the real values.
Here's what I've done:
for any $ 0<t $
$ \intop_{0}^{t}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx=\intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx-\intop_{0}^{t}\frac{C}{x+2}dx=\intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx-C\ln\left(t+2\right)+C\ln2  $
Now, we'll calculate the integral $ \intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx $
$ \intop_{0}^{t}\frac{1}{\sqrt{x^{2}+4}}dx=\intop_{0}^{t}\frac{1}{2\sqrt{\left(\frac{x}{2}\right)^{2}+1}}dx $
And let $ \frac{x}{2}=\sinh u $
so $ dx=2\cosh udu $
and we get
$ \intop_{0}^{t}\frac{1}{2\sqrt{\left(\frac{x}{2}\right)^{2}+1}}dx=\intop_{0}^{aricsinh\left(\frac{t}{2}\right)}1du=arcsinh\left(\frac{t}{2}\right)=\ln\left(\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}\right) $
In order to determine if the integal converges, its enough to calculate the limit :
$ \lim_{t\to\infty}\left(\ln\left(\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}\right)-c\ln\left(t+2\right)\right) $
If c=1 then the limit will be 0, and thus the integral converge, because we'll have:
$ \lim_{t\to\infty}\ln\left(\frac{\frac{t}{2}+\sqrt{1+\left(\frac{t}{2}\right)^{2}}}{t+2}\right)=\ln\left(1\right)=0 $
How can I determine for $ C\neq1 $ ?
Is there a way to calculate the limit for $ C\neq 1 $ ?
Thanks in advacnce
 A: $$I(p)=\int_{0}^{p}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx=\sinh ^{-1}\left(\frac{p}{2}\right)-C (\log (p+2)-\log (2))$$ Using Taylor expansion
$$I(p)=(1-C) \log (p)+C \log (2)-\frac{2 C}{p}+O\left(\frac{1}{p^2}\right)$$ Then ... ??
A: Note
\begin{eqnarray}
&&\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\\
&=&\frac{x+2-C\sqrt{x^{2}+4}}{(x+2)\sqrt{x^{2}+4}}\\
&=&\frac{(x+2)^2-C^2(x^{2}+4)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}\\
&=&\frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}
\end{eqnarray}
If $1-C^2>0$ or $|C|<1$, then there is $N>0$ such that, for $x>N$,
$$ \frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}>0 $$
and
$$ (1-C^2)x^2+4x+4(1-C^2)=O(x^2), (x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})=O(x^3)$$
which implies
$$ \frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}=O(\frac1x) $$
and hence
$$ \int_N^\infty\bigg(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\bigg)dx $$
diverges. So does
$$ \int_0^\infty\bigg(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\bigg)dx. $$
If $1-C^2=0$, then $C=\pm1$. For $C=-1$
$$ \frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}=\frac{1}{\sqrt{x^{2}+4}}+\frac{1}{x+2}>\frac{1}{x+2}, $$
and hence the integral diverges.
For For $C=1$
$$ \frac{1}{\sqrt{x^{2}+4}}-\frac{1}{x+2}=\frac{4x}{(x+2)\sqrt{x^{2}+4}(x+2+\sqrt{x^2+4}))}=O(\frac{1}{x^2}) $$
and hence the integral converges.
