An improper integral of type 2 is an integral whose integrand has a discontinuity in the interval
of integration $[a,b]$. This type of integral
may look normal, but it cannot be evaluated using FTC II, which requires a continuous
integrand on $[a,b]$.
Warning: Now that we have
introduced discontinuous integrands, you will need to check
every integrand you work with for any discontinuities on the
interval of integration.
Examples: $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ and $\displaystyle\int_{-1}^1 \frac{dx}{x^2}$ are of type 2, since $\displaystyle\lim_{x\to0}\frac{1}{\sqrt x}$ and $\displaystyle\lim_{x\to0}\frac{1}{x^2}$ do not exist, and $0$ is contained in the intervals $[0,1]$ and $[-1,1]$, respectively.
We evaluate integrals with discontinuous integrands by taking a
limit; the function is continuous as $x$ approaches the
discontinuity, so FTC II will
work.
Example: $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{t \to 0^+} \int_t^1 \frac{dx}{\sqrt{x}}$.If the discontinuity is in the middle of the interval of integration, we need to break the integral at the point of discontinuity into the sum of two integrals and take limits on both integrals. In this case, we have $t$ approaching the discontinuity from inside the interval of integration of each integral.
Example: $\displaystyle\int_{-1}^1 \frac{dx}{x^2} = \int_{-1}^0 \frac{dx}{x^2} + \int_0^1 \frac{dx}{x^2} =\lim_{t \to 0^-} \int_{-1}^t \frac{dx}{x^2}+\lim_{t \to 0^+} \int_t^1 \frac{dx}{x^2}.$
As with infinite interval integrals, the improper integral converges if the corresponding
limit exists, and
diverges if it
doesn't. When we have to break an integral at the point of
discontinuity, the original integral converges only if both pieces
converge.