Improper integrals are integrals where either:

  1. At least one of the bounds are infinite.
  2. A function is being integrated over an interval which contains an infinite discontinuity (i.e. vertical asymptote).

the first type of improper integral

Here’s an example of the first type of improper integral:

$$ \int_0^\infty e^{-x}\,dx $$

What does this even mean? How can we integrate over an infinite interval?

Well, let’s just see what happens if we take the standard integral $\int_0^b e^{-x}\,dx$ and increase the upper bound to larger and larger numbers, letting it approach infinity:

https://chaddypratt.org/math2419/improper-integral

As you can see, as we let the upper bound $b$ approach infinity, the total area under the curve approaches 1. This means that it makes sense to say, in a sense, that the area under the curve from $x=0$ to $x=\infty$ is 1.

More formally, we can find this improper integral by rewriting it as a limit where the upper bound $b$ approaches infinity:

$$ \int_0^\infty e^{-x}\,dx = \lim_{b\to\infty}\int_0^b e^{-x}\,dx = \lim_{b\to\infty}(-e^{-x})\bigg|0^b = \lim{b\to\infty}[-e^{-b} - (-e^{-0})] =\lim_{b\to\infty}[-e^{-b} + 1] = 1 $$

In general:

$$ \int_a^\infty f(x)\,dx = \lim_{b\to \infty}\int_a^b f(x)\,dx $$

We can do something similar for integrals where the lower bound is $-\infty$.

$$ \int_{-\infty}^b f(x)\,dx = \lim_{a\to -\infty}\int_a^b f(x)\,dx $$

What about integrals where both bounds are infinite? In that case, we can use integral properties to split the integral into two separate improper integrals.

$$ \int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^cf(x)\,dx + \int_c^\infty f(x)\,dx $$

Here, $c$ can be any real number. Note that both of the integrals on the right-hand side must be convergent for the left-hand integral to be convergent (I will explain what convergent integrals are soon).

the second type of improper integral