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.