# How can I determine all the values of p where an integral is improper?

I have this integral:

$\int_0^1{\frac{1}{x^p}}\mathrm{d} x$

And the problem is to find all the values of $p$ for which the integral is improper.

According to the calculus text book I have, and WolframMathWorld, "improper integral" is defined as "[...] a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration". However, Wikipedia and Paul's Online Math Notes seem to be saying that the definition of "improper integral" also includes an integral whose integrand contains any discontinuity at all.

For this problem, using the Wolfram definition: the bounds of integration on this integral are not infinite, so we should only need to determine whether or not the integrand "approaches infinity at one or more points in the range of integration".

Immediately I can see that the integrand only approaches infinity when the term in the denominator approaches $0$, because:

$$\lim_{n \to 0^{\pm}} \frac{1}{n} = \pm \infty$$

Now if I set the denominator of the integrand equal to $0$, and solve, that should give me all of the cases where the integral is improper:

$x^p = 0$

$x = 0^{1/p}$

$0^n$ is defined only when $n > 0$, therefore:

$\frac{1}{p} > 0$

$p > 0$

And this makes sense, because if $p > 0$, then $\frac{1}{x^p}$ goes to $\infty$ as $x$ goes to $0$, whereas, if $p < 0$ then the integrand becomes $\frac{1}{x^{-p}} = x^p$, which is defined and finite over the range of integration.

However, my issue at this point is two-fold:

Firstly, I'm not sure about the case where $p = 0$, by the Wikipedia definition $\int_0^1\frac{1}{x^0}\mathrm{d} x$ is improper because the integrand contains a discontinuity at $x = 0$, which would require a limit to evaluate, but by the Wolfram definition, it isn't improper, because the discontinuity is finite. Which definition is the "correct" one?

Secondly, is my method of solving the problem valid? It works out in this case, but with my example limit:

$$\lim_{n \to 0^{\pm}} \frac{1}{n} = \pm \infty$$

The variable $n$ is approaching $0$, it's not equal to $0$, so setting the denominator equal to $0$ and solving seems like the wrong step. Shouldn't I be taking the limit as the denominator approaches $0$? But in this case, the denominator is a function, not a single variable, so how would I do that?

The limit:

$$\lim_{x^p \to 0^{\pm}} x^p$$

Seems like the wrong notation, and even if not, I'm not sure how I'd evaluate/interpret that. Plus, if the Wikipedia definition is the correct one, my method of solving the problem wouldn't have produced the correct answer, because I would have needed to exclude $p = 0$ as well.

In general if you have an integral that looks like:

$\int_0^1 \frac{1}{f(x,p)}\mathrm{d} x$

How can you determine what values of $p$ make the integral improper?

The integrand is a well-behaved function except perhaps at 0. Thus, if you replace

$\int_0^1{\frac{1}{x^p}}\mathrm{d} x$

with

$\int_{\epsilon}^{1}{\frac{1}{x^p}}\mathrm{d} x$,

for some small $\epsilon >0$, then you can safely integrate:

$\int{\frac{1}{x^p}}\mathrm{d} x = \int x^{-p} \mathrm{d}x = \frac{1}{-p+1}x^{-p+1} + C$, except for the special cases $p=0,1$ with which you can deal separately. Now substitute the limits of integration and check when letting $\epsilon \to 0$ results in a blow-up. To be precise, you might want to consider $\epsilon = \frac{1}{n}$ and send $n \to \infty$.

Let me also address your concern about $\int_0^1{\frac{1}{x^0}}\mathrm{d} x$. Recall that $x^0 = 1$ for all $x$ (including $x=0$), so in fact $\frac{1}{x^0} = 1$ and hence there is no discontinuity.

Finally, for a general $f$, there is probably no one recipe, but usually one has to be careful with end-points of integration (e.g. replace $0$ by $\epsilon$ and $1$ by $1-\delta$) and also identify for what $x$ the function $f$ blows up (if any) and whether this translates to the blow-up for the value of the integral.