Can you refresh my memory on how to easily tell an integral like this one is divergent? It has something to do with the denominator being raised to a power, but I don't quite remember it.

$\displaystyle int[2dx/(x-4)^3]$ for $\displaystyle 0<=x<=5$

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- Apr 6th 2010, 12:33 PMMattpdProving an improper integral divergent.
Can you refresh my memory on how to easily tell an integral like this one is divergent? It has something to do with the denominator being raised to a power, but I don't quite remember it.

$\displaystyle int[2dx/(x-4)^3]$ for $\displaystyle 0<=x<=5$ - Apr 6th 2010, 12:44 PMZevWolf
But thats not an improper integral from what I'm seeing.

An improper integral would have -infinity or +infinity in the boundaries.

Edit: it is improper, because of the denominator being 0 at x=4 - hold on, I'll return.

There's a law for p-series, I remember, about 1/x^2 or 1/x^3 being convergent , but not 1/x because its not fast enough, if that is what you mean.

Edit 2: I graphed it, and here's my answer. Yes, it is divergent. Look at the limits.

Limit as you approach 4 from the negative (left side) is -infinity.

Limit as you approach 4 from the positive (right side) is +infinity.

Does that help? - Apr 6th 2010, 12:49 PMMath Major
So we have that

$\displaystyle \int_{0}^{5} \frac{dx}{(x-4)^3} $

Notice that this integral isn't defined at x = 4. So, we're going to need to split this integral up into two parts and take the limit as the integral approaches 4 as both an upper and lower bound.

$\displaystyle \mathop{\lim}\limits_{n \to ^4} ( \int_{0}^{^-n} \frac{dx}{(x-4)^3} + \int_{n^+}^{5} \frac{dx}{(x-4)^3} ) $

Now evaluate the integrals and see if the limit is convergent or divergent. - Apr 6th 2010, 12:50 PMMattpd
so if the denominator hits 0 anywhere in the interval, and the power the denominator is being raised to is greater than 1, it is divergent?

- Apr 6th 2010, 12:55 PMMath Major
It's divergent here because when you split the integral up into pieces and take the limit as it approaches 4 from either side, you end up evaluating a divergent limit.

- Apr 6th 2010, 01:06 PMMattpd
I understand this, but there is a way to tell just from looking at the problem. No need to split it up or take any integral. Maybe it's called the p-test, but I forget. I think it's what ZevWolf spoke of.

- Apr 6th 2010, 01:10 PMMath Major
In principle, it's very easy to tell here that the integral will be divergent. Notice that the integral is undefined at x = 4. By considering the form of the integral, you can easily guess what it will look like in general, if not in particular. The denominator will still be undefined at x = 4, so it will be divergent.

- Apr 7th 2010, 12:11 AMhollywood
Other posters have covered this, but just to be clear: the integral is "improper" because the integrand goes to infinity at x=4. Other cases of improper integrals are if the limits of integration are negative infinity or positive infinity. An improper integral can either converge or diverge depending on whether the defining limit or limits exist or not.

In this case, the integral is defined by:

$\displaystyle \int_{0}^{5}\frac{dx}{(x-4)^3}=\lim_{x \to 4-}\int_{0}^{x}\frac{dt}{(t-4)^3}+\lim_{x \to 4+}\int_{x}^{5}\frac{dt}{(t-4)^3}$

In order for the integral to converge, both limits have to exist on their own. You get into trouble if you try to combine them before taking the limit.

In this case, both limits go to infinity, so the integral diverges:

$\displaystyle \lim_{x \to 4-}\int_{0}^{x}\frac{dt}{(t-4)^3}=\lim_{x \to 4-}\left[\frac{-1}{2(t-4)^2}\right]_{0}^{x}=\lim_{x \to 4-}\left(\frac{-1}{2(x-4)^2}-\frac{-1}{2(0-4)^2}\right)$

and the first term goes to infinity.

Since we know that the integral of $\displaystyle x^n\text{ is }\frac{x^{n+1}}{n+1}$ except for n=-1, when it is ln |x|, we know that integrals with upper limit or lower limit equal to zero diverge for $\displaystyle n\le-1$ and converge for $\displaystyle n>-1$. This is the p-test you spoke of. So we could have saved some time by noting that we have the case n=-3 and immediately concluded that the integral diverges.

- Hollywood