How to calculate the improper integral $\int_0^\infty\left(\frac{1}{\sqrt{x^2+4}}-\frac{P}{x+2}\right)dx$

Question

This is the first time I've seen a problem like this. I have no idea what to do. Detailed guidance would be of great help.

For which values of P does the integral converge?

$$\int_0^\infty\left(\dfrac{1}{\sqrt{x^2+4}}-\dfrac{P}{x+2}\right)dx$$

Thank You!

Answer 1

Let our function be $f(x)$. The function $f(x)$ behaves nicely in the interval $[0,1]$, so it is enough to find $p$ such that $\int_1^\infty f(x)\,dx$ converges.

If $p\le 0$, then Comparison shows divergence. So we can assume $p\gt 0$.

Rewrite $f(x)$ as $\frac{x+2-p\sqrt{x^2+4}}{(x+2)\sqrt{x^2+4}}$, and then rationalize the numerator by multiplying top and bottom by $x+2+p\sqrt{x^2+4}$. We get $$f(x)=\frac{x^2(1-p^2) +4x+4(1-p^2)}{(x+2)\sqrt{x^2+4}\left(x+2+\sqrt{x^2+4}\right)}.$$

If $p=1$, then $f(x)\le \frac{4x}{2x^3}=\frac{2}{x^2}$. Since $\int_1^\infty \frac{2}{x^2}\,dx$ converges, so does $\int_1^\infty f(x)\,dx$.

If $0\lt p\lt 1$ or $p\gt 1$, then our integral diverges. There are two cases to consider.

Suppose that $0\lt p\lt 1$. Let $g(x)=\frac{1}{x}$. One can show that $$\lim_{x\to\infty} \frac{f(x)}{g(x)}=\frac{1-p^2}{2}.$$ Since $\int_1^\infty g(x)\,dx$ diverges, so does $\int_1^\infty f(x)\,dx$. The argument for $p\gt 1$ is similar.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{\Lambda > 0}$: \begin{align} &\color{#c00000}{% \int_{0}^{2\Lambda}\pars{{1 \over \root{x^{2} + 4}} - {P \over x + 2}}\,\dd x} ={\rm arcsinh}\pars{\Lambda} - P\ln\pars{\Lambda + 1} \\[3mm]&=\ln\pars{\Lambda + \root{\Lambda^{2} + 1}} - P\ln\pars{\Lambda + 1} =\ln\pars{\Lambda + \root{\Lambda^{2} + 1} \over \bracks{\Lambda + 1}^{P}} \end{align}

When $\ds{\Lambda \gg 1}$, it's clear that:

1. $\ds{\large P \not= 1}$: The integral diverges as $\ds{\ln\pars{2\Lambda^{1 - P}}}$.
2. $\ds{\large P = 1}$: The integral converges to $\ds{\ln\pars{2}}$.