I gave a rather unclear explanation of the following in a lecture a couple of days ago and promised to (at least attempt to) clarify a bit.

You know that:

$\int_0^{10}\frac{1}{x} dx$

is divergent and gives an answer of $\infty$. If you have an integral of the form:

$\int_3^{10}\frac{1}{x-3}$

as we had in class, you don’t need to calculate this separately if you already know about the behaviour of $1/x$ about $x=0$ as they are exactly the same thing, just shifted (at least their convergence properties are the same).

The take-home message is that if you have:

$\int_a^b \frac{c}{(x-a)^p}dx$

and you already know the behaviour of:

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

then the integral you are interested in will have the same convergence or divergence behaviour.

 How clear is this post?