TITLE: Why can't you integrate over a discontinuity?
QUESTION [5 upvotes]: I'm asked to investigate this integral,
$$\int_0^3 \frac{1}{(1-x)^2}\,\mathrm{d}x$$
I get -3/2, but the integrand has a discontinuity at $x=1$, so I can't just integrate it. What is the answer? And why do things go wrong when integrating over a discontinuity?
Also what about $$\int_{-1}^{1} \frac{1}{x}\,\mathrm{d}x$$
When evaluated it gives 0, but this is correct since the areas cancel.

REPLY [5 votes]: It is not so much the discontinuity of the integrand that is of issue as the discontinuity of the naïvely computed antiderivative, i.e. $F(x)=1/(1-x)$. The theorem that $$\int_a^b F'(x)\,dx=F(b)-F(a)$$
requires at the very least that $F$ be continuous on $[a,b]$, and that is not the case here.
The second integral you mention can be fruitfully considered to be zero by its
Cauchy principal value. Whether that is appropriate or not depends on the intended use for the integral.