The Comparison Test for Improper Integral Convergence/Divergence

Suppose we are interested in determining if an improper integral converges or diverges as opposed to simply evaluating the integral. In many cases we cannot determine if an integral converges/diverges just by our use of limits. However, the following theorem will allow us to determine if a curve converges/diverges without actually evaluating the integral.

Theorem 1 (The Comparison Test for Integral Convergence/Divergence): Suppose that $0 ≤ g(x) ≤ f(x)$. If $\int_a^{\infty} f(x) \: dx$ also converges, then $\int_a^{\infty} g(x) \: dx$ converges. Furthermore, if $\int_a^{\infty} g(x) \: dx$ diverges, then $\int_a^{\infty} f(x) \: dx$ also diverges.

• Proof: Let $0 ≤ g(x) ≤ f(x)$ and suppose that $\int_a^{\infty} f(x) \: dx$ converges. Then $\int_a^{\infty} f(x) \: dx = A ≥ 0$. Therefore $0 ≤ \int_b^{\infty} g(x) \: dx ≤ A$ and so $\int_b^{\infty} g(x)$ converges.

• The second part of theorem 1 is the contrapositive of the proof given above. $\blacksquare$

We note that if we have a function $g(x)$ and can find a function $f(x)$ such that $0 ≤ g(x) ≤ f(x)$ and that $f(x)$ converges on the interval $[0, \infty )$, then it must follow that $g(x)$ also converges. Similarly, if we have a function $f(x)$ and can find a function $g(x)$ where $g(x)$ diverges on the same interval, then it follows that $f(x)$ must ALSO diverge. The following illustration describes this property:

We note that this comparison test must always hold true since the area of $g(x)$ is always contained in the area of $f(x)$. Hence if $f(x)$ is finite, then so is $g(x)$. Similarly if $g(x)$ is infinite, then so is $f(x)$.

Example 1

Using the Comparison Test, determine if $\int_1^{\infty} e^{-x^2} \: dx$ converges or diverges.

Recall that to use the comparison test to determine if a function converges, we must find a function, let's call it $f$ such that $0 ≤ e^{-x^2} ≤ f(x)$ and that also converges. We will only look at $x ≥ 1$.

Suppose that we choose $f(x) = e^{-x}$. We note that $0 ≤ e^{-x^2} ≤ e^{-x}$ for $x ≥ 1$. Note that the only difference between these two functions is their exponents. We note that the exponents are both negative, so the function with the larger negative exponent will be smaller. Furthermore, we note that $x^2 ≥ x$ for $x ≥ 1$, and hence it follows that $0 ≤ e^{-x^2} ≤ e^{-x}$ for $x ≥ 1$. Now let's determine if $f(x)$ converges or not:

(1)

\begin{align} \int_1^{\infty} e^{-x} \: dx = \lim_{b \to \infty} \int_1^b e^{-x} \: dx \\ \int_1^{\infty} e^{-x} \: dx = \lim_{b \to \infty} \left ( -e^{-x} \right )\bigg| _1^b \\ \int_1^{\infty} e^{-x} \: dx = \lim_{b \to \infty} \left ( -e^{-b} + e^{-1} \right ) \\ \int_1^{\infty} e^{-x} \: dx = \lim_{b \to \infty} \left ( -\frac{1}{e^{b}} + \frac{1}{e} \right ) \\ \int_1^{\infty} e^{-x} \: dx = \frac{1}{e} \end{align}

Hence $\int_1^{\infty} e^{-x} \: dx$ converges. Since $0 ≤ e^{-x^2} ≤ e^{-x}$ for $x ≥ 1$, it follows that $\int_1^{\infty} e^{-x^2} \: dx$ must also converge.