Improper integrals are improper for one of two reasons:

1. The integrand has a singularity within the range of integration or at an endpoint.

2. The range of integration is infinite in length.

Suppose that integral I has only one singularity or impropriety.

In the first case, we can make I proper by excluding all points within distance d of the singularity from the range of integration. This will produce a proper integral I(d).  

In the second case, we can integrate to R instead of infinity, obtaining a proper integral I(R). I will exist if exist, or exists.

When does this happen?

If f(x) behaves like a power at the singular point a, ie like , then for the integral behaves like .

Thus, if a is finite there is convergence for p > -1, ergence for p <  -1.

If the integral is improper because it extends to infinity, then it will converge for p < -1.

What happens for p = 1? The integral of the absolute value of the integrand will erge; the integral of f may exist in some sense through cancellations of positive and negative contributions.