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:
is divergent and gives an answer of . If you have an integral of the form:
as we had in class, you don’t need to calculate this separately if you already know about the behaviour of about 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:
and you already know the behaviour of:
then the integral you are interested in will have the same convergence or divergence behaviour.