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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?

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  • $\begingroup$ Why would the integral diverge on $(0,1)$ for example in the case $a=0$, $b=2$? $\endgroup$ – Troposphere May 6 at 2:05
  • $\begingroup$ @Troposphere , thx for comment, $\int_{0}^{1}x^{-3} = -{1 \over 2}(1^{-2} - 0^{-2}) = \infty$ $\endgroup$ – ZKK May 6 at 2:13
  • $\begingroup$ What does $\int_0^1 x^{-3}dx$ have to do with whether $\int_0^1\frac{dx}{(4+9x)^3}$ has a value or not? The latter is not even improper; the integrand is always between $0$ and $1/64$. $\endgroup$ – Troposphere May 6 at 2:14
  • $\begingroup$ @Troposphere my thought's like: 1. by the limit comparison test, $\int_{0}^{\infty} dx/(4+9x)^3$ and $\int_{0}^{\infty}dx/x^3$ both converge or diverge; 2. $\int_{0}^{1}dx/x^3$ diverges, thus $\int_{0}^{\infty}dx/x^3$ diverges; 3. thus, $\int_{0}^{\infty} dx/(4+9x)^3$ diverges; $\endgroup$ – ZKK May 6 at 2:47
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    $\begingroup$ I suspect your mistake is that you have overlooked that $x^{-3}$ lacks a limit for $x\to 0$ whereas $(4+9x)^{-3}$ has one. The "limit comparison test" I could find with Google (e.g. at math.fel.cvut.cz/mt/txtd/4/txe3da4c.htm) requires both functions to be defined at the lower endpoint; $x^{-3}$ is not defined at $0$. $\endgroup$ – Troposphere May 6 at 3:09
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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 $a + b > 0$.

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  • $\begingroup$ "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 $a < 1$." Sorry I don't understand this step, how to define "small"? and why the simulation abandons the part $x^{b+1}$? $\endgroup$ – ZKK May 6 at 3:18
  • $\begingroup$ @ZKK See the edit. $\endgroup$ – Mark Saving May 6 at 3:22
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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 $1-a>0$ $$\Rightarrow\,b>-a>-1$$

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