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In order to prove that $\int_{0}^{0.5}\frac{1}{x^2-1}$ converges I compared it to $\int_{0}^{0.5}\frac{1}{x}$ which converges, by checking that $\lim \frac{\frac{1}{x^2-1}}{\frac{1}{x}}=0$ when $x \to 0$. Then I went to Wolfram Alpha and tried to check $\int_{0}^{0.5}\frac{1}{x^2-0.1}$ (It is suppose to converges by the same test) but it said that it diverges, while $\int_{0}^{0.5}\frac{1}{x^2-0.3}$ doesn't. (with $0.2$ it just couldn't compute). What's really going on there between $0.1$ to $0.3$? Is W.A wrong?

Edit: Sorry! $\int_{0}^{0.5}\frac{1}{x}$ obviously doesn't converge. So, in addition to the rest of the question, how can I prove my original integral does converge?

Thank you very much.

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Short answer to the edited question: Because $x^2$ ranges from $0$ to $0.25$, and $0.1$ is in this range (making the denominator $0$) whereas $0.3$ is outside of this range. –  Eric Naslund Feb 2 '12 at 14:23
    
It should also be pointed out that there are missing $dx$ terms in most of your posts (at least when they originally are written). This is a bad habit to start! –  JavaMan Feb 2 '12 at 17:58
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2 Answers

up vote 3 down vote accepted

Between $x=0$ and $x=0.5$, the function $\frac{1}{x^2-1}$ is perfectly respectable! Note that the denominator is never $0$ in our interval. The largest absolute value is reached at $x=0.5$. So your function has no issues, it is continuous on a closed interval. For the problem you were initially considering, we are finished.

But the question you were led to ask is more interesting, and shows a good effort to understand the situation.

Look first at $\frac{1}{x^2-0.3}$. The denominator is $0$ at $x=\pm\sqrt{0.3}$. The positive root is roughly $0.547722$, outside our interval, though not by much. Thus the function $\frac{1}{x^2-0.3}$ is well-behaved in the interval $[0,0.5]$.

Look now at $\frac{1}{x^2-0.1}$. The denominator is $0$ at $x=\pm\sqrt{0.1}$. The positive root is about $0.3162278$, and this is inside our interval. So our function blows up inside our interval, and there may be a problem. Indeed there is.

You know that a function can blow up, but despite that the integral converges. A standard example is $\int_0^1 \frac{dx}{\sqrt{x}}$. We will show that $\int_0^{0.5}\frac{dx}{x^2-0.1}$ diverges.

As mentioned above, there is potential trouble at $\sqrt{0.1}$. To make typing easier, let $a=\sqrt{0.1}$. Our function is not defined at $x=a$, and blows up near $x=a$. Recall that $a$ is in our interval. When we are dealing with a singularity inside our interval, it is useful to break up the interval into two integrals, in this case from $0$ to $a$ and from $a$ to $0.5$.

We will show that $\int_0^a \frac{dx}{x^2-0.1}$ diverges. (The integral from $a$ to $0.5$ also does, but showing that one of the integrals is bad is enough.)

So we are looking at the integral $\int_0^a\frac{dx}{x^2-a^2}$. For no good reason, except for a preference for the positive, we look instead at $$\int_0^a \frac{dx}{a^2-x^2}.$$ Make the change of variable $w=a-x$. Note that $a^2-x^2=(a-x)(a+x)=w(2a-w)$. Quickly we arrive at $$\int_{w=0}^a \frac{dw}{w(2a-w)}.$$ This integral diverges, by comparison with $\int_0^a\frac{dw}{w}$, which, as pointed out by anonymous, diverges.

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Thanks a lot! very helpful and clear. –  Jozef Feb 2 '12 at 15:28
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You have a mistake. The integral $\int_{0}^{0.5} dx/x$ does not converge! in fact, we can easily calculate it, as the antiderivative of $1/x$ is $lnx$, and $\lim_{x\to 0^+} ln(x) = -\infty$.

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Right! so how can I prove that my original integral does converge? –  Jozef Feb 2 '12 at 14:12
    
Well, your integral is actually a definite integral. The function $\frac{1}{x^2-1}$ is continuous on the interval $[0,1/2]$, so there is no question of convergence at all. –  anonymous Feb 2 '12 at 14:13
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