Your working is perfectly correct, the only incorrect logic is the following:
I am clearly wrong since Wolfram|Alpha disagrees with me
To compute the integral Wolfram|Alpha has used a numerical method to estimate the value and the method it is using is not sophisticated enough to handle this integral near the asymptote.
This is not uncommon. I can't remember specific instances off the top of my head, but several times I've been able to work out simpler expressions than the forms returned in WA. Humans are still usually better mathematicians than computers (13/02/12) .
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.
Best Answer
As @eyeballfrog notes in the comments, you may find it helpful to think of the intuition for $$\sum_{n=1}^\infty 2^{-n}=1.$$ Visually, this says that if you fill a glass half full, then fill half of the empty space remaining, then fill half of the empty space remaining, etc. you will approach filling one glass. Despite infinitely many fills, we have a finite volume in the limit because intuitively each fill introduces sufficiently less volume than the preceding fill (with vanishing volume in the limit).
A continuous analogue of the above series is the improper integral
$$\int_0^\infty e^{-x}dx=1.$$
Again, as with cup filling, though integrating over an infinite interval, $e^{-x}$ vanishes sufficiently quickly in the limit for the integral to be convergent.
The above improper integral has an integrand with a horizontal asymptote. An improper integral whose integrand has a vertical asymptote may be thought of much the same way; just think of the integrand as a function with a horizontal asymptote that has been rotated 90 degrees.