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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?


Thank You!

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You calculate improper integrals the way you would any definite integral, with the modification that you need to take limits where appropriate. Find the antiderivative of each term. The lower limit of integration is zero and the antiderivatives are defined there. For the upper limit, set up antiderivatives with a dummy variable, $ \ F(t) \ $ and find the limit of that antiderivative function as $ \ t \ $ goes to infinity. – RecklessReckoner May 18 '14 at 18:19
On a related note, the first term by itself will not produce a finite value, but both terms together will for certain values of that unspecified constant $ \ P \ $ . Is that something you're supposed to figure out values for that make the improper integral produce a finite value? – RecklessReckoner May 18 '14 at 18:24
RecklessReckoner, thaks for your answer. The answer to your question is yes. I have to find which values of P the integral converges. – luis May 18 '14 at 18:44
I believe you will want to write out the antiderivative of the complete integrand first. It looks like there is a specific value of $ \ P \ $ that will cause the divergent terms to cancel out, so that the improper integral will converge (give a finite value). – RecklessReckoner May 18 '14 at 19:05

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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 $\ds{\ln\pars{2}}$.
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Your tex source seem a little bit interesting... :-) – peterh May 19 '14 at 7:36
@PeterHorvath Thanks a lot. I always try to be as clear as possible. – Felix Marin May 19 '14 at 7:49

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.

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