What is all possible value of a,b $\in\mathbb{R}$ so that the following integral converges:$\int_{0}^{+\infty}{dx\over x^a(4+9x)^{b+1}}$ What is all possible value of a,b $\in\mathbb{R}$ so that the following integral converges:
$\int_{0}^{+\infty}{dx\over x^a\space(4+9x)^{b+1}}$
I want to use the limit comparison test. Since $\lim_{x\to \infty}{{x^a\space (4+9x)^{b+1}}\over{9^{b+1} 
\space\space  x^{a+b+1}}} = constant$, then the improper integral $\int_{0}^{\infty}{dx\over{9^{b+1} 
\space\space  x^{a+b+1}}}$   and $\int_{0}^{\infty}{dx\over x^a\space(4+9x)^{b+1}}$ should both converge or diverge.
Then we consider ${1 \over {x^{a+b+1}}}\space $
if $a+b+1 = 1$ then the improper integral $\int_{0}^{\infty}{1 \over {x^{a+b+1}}}\space $diverges;
if $a+b+1<1$ then the improper integral diverges on ($1$,$\infty$);
if $a+b+1>1$ then the improper integral diverges on ($0$,$1$)
So it seems that the applicable set of (a,b) is empty. Is that right?
 A: The two improprieties we must consider are $0$ and $\infty$.
For small $x$, we have $x^a (4 + 9x)^{b + 1} \sim x^a 4^{b + 1}$, and we know that $\int\limits_0^1 \frac{1}{x^a} dx$ converges iff $a < 1$. So we require that $a < 1$.
Edit: to be precise, $\lim\limits_{x \to 0} \frac{x^a (4 + 9x)^{b + 1}}{x^a 4^{b + 1}} = 1$. Thus, $\int\limits_0^1 \frac{1}{x^a (4 + 9x)^{b + 1}} dx$ converges iff $\int\limits_0^1 \frac{1}{x^a} dx$ converges, which occurs iff $a < 1$.
For large $x$, we have $x^a (4 + 9x)^{b + 1} \sim x^{a + b + 1} 9^{b + 1}$, and we know that $\int\limits_1^\infty \frac{1}{x^n} dx$ converges iff $n > 1$. So we require that $a + b + 1 > 1$: that is, $a + b > 0$.
So the requirements are $a < 1$ and $a + b > 0$.
A: To long for a comment.
Integral can be evaluated in the closed form:
$$I(a,b)=\int_{0}^{+\infty}{dx\over x^a(4+9x)^{b+1}}=\frac{1}{4^{b+1}}\int_{0}^{+\infty}{dx\over x^a(1+\frac{9}{4}x)^{b+1}}=\frac{9^{a-1}}{4^{a+b}}\int_{0}^{+\infty}{dt\over t^a(1+t)^{b+1}}$$
Making change $x=\frac{1}{1+t}$
$$I(a,b)=\frac{9^{a-1}}{4^{a+b}}\int_{0}^{1}x^{a+b-1}(1-x)^{-a}dx=\frac{9^{a-1}}{4^{a+b}}B(a+b;1-a)=\frac{9^{a-1}}{4^{a+b}}\frac{\Gamma(a+b)\Gamma(1-a)}{\Gamma(b+1)}$$
Beta-function (in the form of this integral) is finite and defined for $a+b>0$ and $1-a>0$ $$\Rightarrow\,b>-a>-1$$
