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$$\int_0^2 \dfrac{\mathrm dx}{\sqrt{x}(x-1)}$$

I want to determine whether this integral converges or diverges. Now usually problems like these are easy, but this one is kind of tricky since it is discontinuous at both 0 and 1. Whereas with one situations where the integral is only discontinuous at 1 value, I could just set up 2 integrals and use the "lim of t" method, but here I cant set up two integrals. The only way that two integrals could be set up is if the first one is from 0 to 1 and the second one is from 1 to 2, but the first one is discontinuous at both values, so not sure what the work around is.

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You can split it into three integrals: $0$ to $0.5$, $0.5$ to $1$, $1$ to $2$. –  Timothy Wagner Dec 7 '10 at 22:52
    
Oh I can use decimals? Wasnt sure about that.. –  maq Dec 7 '10 at 22:53
    
@fprime: Yes you can use any real number as limit of integration. –  Timothy Wagner Dec 7 '10 at 23:01

3 Answers 3

up vote 6 down vote accepted

You can break up the integral at any point or points you like. In this case, you could break it up into three integrals: pick a point $c$ strictly between $0$ and $1$, and consider: \begin{align*} \int_0^2 \frac{dx}{\sqrt{x}(x-1)} &= \int_0^c\frac{dx}{\sqrt{x}(x-1)}+\int_c^1\frac{dx}{\sqrt{x}(x-1)} + \int_1^2\frac{dx}{\sqrt{x}(x-1)}\\ &= \lim_{a\to 0^+}\int_a^c \frac{dx}{\sqrt{x}(x-1)} + \lim_{b\to 1^-}\int_c^b\frac{dx}{\sqrt{x}(x-1)} + \lim_{d\to 1^+}\int_d^2\frac{dx}{\sqrt{x}(x-1)}. \end{align*} The original improper integral exists if and only if each of the three improper integrals exist.

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Sorry for the late comment, but when you say 'exists', do you mean converges? –  maq Dec 9 '10 at 22:49
    
Actually nm, I verified it in my book. If it exists, then it converges. –  maq Dec 9 '10 at 22:50
    
@f-Prime: Yes: I should hae said "three limits exist". Technically, improper integrals converge or diverge, limits exist or don't exist. –  Arturo Magidin Dec 10 '10 at 2:46

Consider the change of variables $t=\sqrt x$. It transforms the interval $[0,2]$ into $[0,\sqrt 2]$ and removes one of the problem zeroes. It is easy to see from the resulting expression that the integral diverges.

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$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$Try to solve the integral. You'll find whether it converges. For example, $\ds{\int_{\epsilon}{\dd x \over \sqrt{x\,}\,\pars{1 - x}}}$ with $\epsilon > 0$. Later on, you can study the limit $\ds{\epsilon \to 0^{+}}$. In this way, you "kill two birds with a one shot".

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