Type 2 - Improper Integrals with Discontinuous Integrands

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:  $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{t \to 0^+} \int_t^1 \frac{dx}{\sqrt{x}}$.

Example:  $\displaystyle\int_{-1}^1 \frac{dx}{x^2} = \int_{-1}^0 \frac{dx}{x^2} + \int_0^1 \frac{dx}{x^2} =\lim_{t \to 0^-} \int_{-1}^t \frac{dx}{x^2}+\lim_{t \to 0^+} \int_t^1 \frac{dx}{x^2}.$

On Convergence

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.