Converging integrals Given this integral:
$$\int_2^∞ \frac {1}{({x-2)}^{2a+b}{(2x+5)}^b} \, dx$$
How do you find out when the integral converges (i.e. what limitations must be placed on a and b for the integral to converge)?
I understand that you analyze how the integral behaves at the integration limits. So you take the limits of the integrand as it approaches the integration limits. So:
$$\frac {1}{({x-2)}^{2a+b}{9}^b}$$ when x = 2. But from this, how do you figure out whether 2a+b should be greater than, equal to, and/or less than 1?
Similarly we get:
$$\frac {1}{{x}^{2a+b}{(2x)}^b}$$ which is equal to $$\frac {1}{{x}^{2a+2b}{2}^b}$$
But again, how does this help show you the circumstances when convergence occurs?
Thanks!
 A: $\int_0^1 x^p\ dx$ converges iff $p > -1$, and $\int_1^\infty x^p\ dx$ converges iff $p < -1$.  So you want $2a+b < 1$ for convergence as $x \to 2$, and
$2a+2b>1$ for convergence as $x \to +\infty$.
A: HINT: Try bounding the integrand from below and above and use the fact that:
$\int_1^\infty \frac 1 {x^p} \, dx $ converges for $p>1$ and
$\int_0^1\frac 1 {x^p} \, dx $ converges for $p<1$
Also changing variables might help.
spoiler:

my guess is that $2a+b=0$ and $b>1$

A: Suggestion: it may be easier to make the change of variables $t=x-2$ and split the new
integral into the following two 
\begin{eqnarray*}
I &=&\int_{2}^{\infty }\frac{1}{(x-2)^{2a+b}(2x+5)^{b}}\,dx=\int_{0}^{1}%
\frac{1}{t^{2a+b}\left( 9+2t\right) ^{b}}\,dt+\int_{1}^{\infty }\frac{1}{%
t^{2a+b}\left( 9+2t\right) ^{b}}\,dt \\
&=&I_{1}+I_{2}.
\end{eqnarray*}
We can use the limit comparison test twice, noticing that the second
integral ($I_{2}$) behaves like
\begin{equation*}
J_{2}=\int_{1}^{\infty }\frac{1}{t^{2a+2b}}\,dt=\int_{1}^{\infty }\frac{1}{
t^{p}}\,dt,\qquad p=2a+2b,
\end{equation*}
as $t\rightarrow \infty $, and the first one ($I_{1}$) like 
\begin{equation*}
J_{1}=\int_{0}^{1}\frac{1}{t^{2a+b}}\,dt=\int_{0}^{1}\frac{1}{t^{q}}
\,dt,\qquad q=2a+b,
\end{equation*}
as $t\rightarrow 0^{+}$. For which values of $p=2a+2b$ and $q=2a+b$ do $J_2$ and $J_1$ converge? 
A: $\newcommand{\+}{^{\dagger}}%
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$$
x \sim 2\quad \imp \quad 2a + b > -1\,,\qquad x \to \infty\quad\imp\quad -2a - 2b + 1 <0
$$
$$ 2a + b > -1\,,\qquad a + b > \half$$
