Computing Improper Integrals with an Undefined Endpoint

Previously, we found out that unbounded regions can, in fact, have finite area! So, a natural next step would be to try and calculate that area. This leads us into the notion of improper integrals.

There are two different kinds of improper integrals: in this section, we deal with integrals in which the integrand is undefined at one or both of the limits of integration. Geometrically, this involves regions that are unbounded vertically.

Let's start with an example, and calculate the integral $$\int_1^2 \frac{1}{\sqrt{2-x}}\, dx.$$ We might be tempted to solve this the ``usual'' way: $$\int_1^2 \frac{1}{\sqrt{2-x}}\, dx=-2\sqrt{2-x}\bigg|_1^2=-2(\sqrt{2-2}-\sqrt{2-1})=2.$$ But wait, there's something subtly wrong with what we just did! Do you see what that could be?

Let's first graph the function in the integrand:

From this picture, we can clearly see that the function ``explodes'' to infinity at $x=2$, and is thus discontinuous at $x=2$. So, this means that we can't directly compute the integral as we just did! So, let's try to see whether we can solve for the integral another way.

In this example, we know that the integrand is continuous for all $t<2$, which means that the new integral $$I(t) = \int_1^t \frac{1}{\sqrt{2-x}}\, dx$$ is well-defined. So, we can think of the value of the original integral $$\int_1^2 \frac{1}{\sqrt{2-x}}\, dx$$ as the value that $I(t)$ approaches as $t\to 2$ from the left. In other words, we can say that $$\int_1^2 \frac{1}{\sqrt{2-x}}\, dx=\lim_{t\to 2^-} \int_1^t \frac{1}{\sqrt{2-x}}\, dx.$$ Now, we can finally solve the problem formally: $$\int_1^2 \frac{1}{\sqrt{2-x}}\, dx=\lim_{t\to 2^-} \int_1^t \frac{1}{\sqrt{2-x}}\, dx=\lim_{t\to 2^-} -2(\sqrt{2-x})\bigg|_1^t$$ $$=\lim_{t\to 2^-} -2(\sqrt{2-t}-\sqrt{2-1})=\lim_{t\to 2^-} (2-2\sqrt{2-t})=2.$$ So, the area of the unbounded region shown above has area $2$.

We can generalize what we just did by stating the following:

1. If $f(x)$ is continuous on $[a,b)$ and not on $x=b$, then $$\int_a^b f(x)\, dx=\lim_{t\to b^-} \int_a^t f(x)\, dx.$$ Similarly, if $f(x)$ is continuous on $(a,b]$ and not on $x=a$, then $$\int_a^b f(x)\, dx=\lim_{t\to a^+} \int_t^b f(x)\, dx.$$ For both cases, the equality holds only if the limit exists and is finite. In that case, we say that the integral converges; otherwise, we say that it diverges.

2. If $f(x)$ is continuous on $[a,b]$ except at $x=c$, where $a<c<b$, then $$\int_a^b f(x)\, dx=\int_a^c f(x)\, dx+\int_c^b f(x)\, dx,$$ as long as both integrals on the right side converge. If either diverge, then so does the integral on the left.

3. If $f(x)$ is continuous on $(a,b)$ but not at $x=a$ and $x=b$, then for any $c$ between $a$ and $b$ we have $$\int_a^b f(x)\, dx=\int_a^c f(x)\, dx+\int_c^b f(x)\, dx,$$ as long as both integrals on the right converge. If either diverge, then so does the integral on the left.

Now that we have an idea of how to deal with integrals that are discontinuous at a point, let's try a few more examples:

First, let's calculate the integral $$\int_0^1 \frac{1}{x^2}\, dx.$$ Since $\frac{1}{x^2}$ is discontinuous at $x=0$ but continuous for all $t>0$, we have $$\int_0^1 \frac{1}{x^2}\, dx=\lim_{t\to 0^+} \int_t^1 \frac{1}{x^2}\, dx=\lim_{t\to 0^+} -\frac{1}{x}\bigg|_t^1\\=\lim_{t\to 0^+} -\left(\frac{1}{1}-\frac{1}{t}\right)=\lim_{t\to 0^+} \left(\frac{1}{t}-1\right)=\infty.$$ What does that mean? Well, let's graph the integrand:

Since the value of the integral is infinity, the area of this unbounded region is not finite! This shows you that not all unbounded regions have infinite area, but some still do!

Now, you try!

If we look at the function $y=\frac{1}{x}$, we see that this is continuous on $[-2,2]$ except at $x=0$. So, we have to split the integral into the following: $$\int_{-2}^2 \frac{1}{x}\, dx=\int_{-2}^0 \frac{1}{x}\, dx+\int_0^2 \frac{1}{x}\, dx.$$ Let's try to solve the right-most integral: $$\int_0^2 \frac{1}{x}\, dx=\lim_{t\to 0^+} \int_t^2 \frac{1}{x}\, dx=\lim_{t\to 0^+} \ln{x}\bigg|_t^2=\lim_{t\to 0^+} (\ln{2}-\ln{t})=\infty.$$ Since this integral diverges, so does the original integral.

Try one more problem!

Since $y\sec^2{x}$ is undefined at $x=\frac{\pi}{2}$, we can write the integral as the following: $$\int_0^{\frac{\pi}{2}} \sec^2{x}\, dx=\lim_{t\to \frac{\pi}{2}^-} \int_0^t \sec^2{x}\, dx=\lim_{t\to \frac{\pi}{2}^-} \tan{x}\bigg|_0^t\\=\lim_{t\to \frac{\pi}{2}^-} (\tan{t}-\tan{0})=\lim_{t\to \frac{\pi}{2}^-} \tan{t}=\infty.$$ So again, the integral diverges.