Extension: Comparison Test for Improper Integrals

In a previous section, we learned about how to find whether integrals of the form $$\int_a^{\infty} f(x)\, dx$$ converge or diverge. Now, let's apply what we've learned so far to more complicated improper integrals.

For example, consider the integral $$\int_1^{\infty} \frac{\sin^2{x}}{x^2}\, dx.$$ This integral seems too complicated for us to compute. However, we can find a relatively easy way to test whether it converges or diverges.

To do so, we will need to find an integral that either has a larger integrand and converges, or has a smaller integrand and diverges.

So, let's focus on the integrand $$\frac{\sin^2{x}}{x^2}.$$

Great! Now that we have a bound for $\sin^2{x}$, we can write $$\frac{\sin^2{x}}{x^2}\le \frac{1}{x^2}.$$ Okay, but how do the integrals $$\int_1^{\infty} \frac{\sin^2{x}}{x^2}\, dx \text{ and } \int_1^{\infty} \frac{1}{x^2}\, dx$$ relate?

To figure out how they relate, let's graph the two integrands:

Since the graph of $f(x)=\frac{1}{x^2}$ always lies above the graph of $g(x)=\frac{\sin^2{x}}{x^2}$, the area between the $x$-axis, the line $x=1$, and the graph of $f(x)$ will always be at least the area between the $x$-axis, the line $x=1$, and the graph of $g(x)$. Therefore, it follows that $$\int_1^{\infty} \frac{\sin^2{x}}{x^2}\, dx \le \int_1^{\infty} \frac{1}{x^2}\, dx.$$ And since we know that the right side is finite, the left side must also be finite. Hence, the original integral converges.

We can generalize what we just discovered into the following results:

Suppose $f(x)$ and $g(x)$ are functions such that $f(x)\ge g(x)\ge 0$ on the interval $[a,\infty)$. Then

1. If $\int_a^{\infty} f(x)$ converges, then $\int_a^{\infty} g(x)$ also converges.

2. If $\int_a^{\infty} g(x)$ diverges, then $\int_a^{\infty} f(x)$ also diverges.

So now let's look at a few more practice problems: Suppose we want to determine whether the integral $$\int_0^{\infty} \frac{1}{17+e^x}\, dx$$ converges or diverges. In order to use the fact stated above, we need to find a smaller integrand whose integral diverges, or a larger integrand whose integral converges. This time, we can remember that for $a,b> 0$, $$a\ge b\Longrightarrow \frac{1}{a}\le \frac{1}{b}.$$ So, since $17+e^x\ge e^x$, it follows that $$\frac{1}{17+e^x}\le \frac{1}{e^x}.$$ And since $$\int_0^{\infty} \frac{1}{e^x}\, dx=\lim_{t\to\infty} \int_0^t e^{-x}\, dx=\lim_{t\to\infty} -e^{-x}\bigg|_0^t$$ $$=\lim_{t\to\infty} -(e^{-t}-e^0)=\lim_{t\to\infty} \left(1-\frac{1}{e^t}\right)=1,$$ the integral $$\int_0^{\infty} \frac{1}{e^x}\, dx$$ converges. Therefore, we also get that $$\int_0^{\infty} \frac{1}{17+e^x}\, dx\le \int_0^{\infty} \frac{1}{e^x}\, dx$$ also converges.

Now you try!

For this case, we note that $17$ and $e^x$ are both greater than $0$, and so since $17\ge 17-e^x$, we get that $$\frac{1}{17}\le \frac{1}{17-e^x}.$$ And since $$\int_0^{\infty} \frac{1}{17}\, dx=\lim_{t\to\infty} \int_0^t \frac{1}{17}\, dx=\lim_{t\to\infty} \frac{1}{17}t\bigg|_0^t=\lim_{t\to\infty} \frac{t}{17}=\infty,$$ the integral $$\int_0^{\infty} \frac{1}{17}\, dx$$ diverges. Therefore, we also have that $$\int_0^{\infty} \frac{1}{17-e^x}\, dx$$ also diverges.