Converging integrals

Question

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!

Answer 1

$\int_0^1 x^p\ dx$ 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 $x \to +\infty$.

Answer 2

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 $b>1$

Answer 3

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?

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ x \sim 2\quad \imp \quad 2a + b > -1\,,\qquad x \to \infty\quad\imp\quad -2a - 2b + 1 <0 $$ $$ 2a + b > -1\,,\qquad a + b > \half$$