Comparison Theorem for Integrals

A theorem to easier determine if a Improper Integral converges or diverges.

Theorem

1. Suppose $f,g,h$ are continuous on interval $[a,b]$
2. Consider ${\int }_a^{b}f(x)dx$ is an improper integral
3. Convergent case:
   1. If $\forall x\in [a,b],0\le f(x)\le g(x)$
   2. And, we also know that ${\int }_a^{b}g(x)dx$ converges
   3. Then, ${\int }_a^{b}f(x)dx$ also converges
4. Divergent case:
   1. If $\forall x\in [a,b],0\le h(x)\le f(x)$
   2. And, we know that ${\int }_a^{b}h(x)dx$ diverges
   3. Then, ${\int }_a^{b}f(x)dx$ also diverges

Proofs

• Comparison Theorem Convergent Case Proof

Examples

Example 1

Consider ${\int }_0^{\infty }{e}^{-x^2}dx$. Determine if it converges or diverges

1. Note that $g(x)=e^{-x}$ converges to 0 as ${\lim }_{n\to \infty }{e}^{-n}=0$
2. By union interval property ${\int }_0^{\infty }{e}^{-x^2}dx={\int }_0^1{e}^{-x^2}dx+{\int }_1^{\infty }{e}^{-x^2}dx$
3. Note that on $x\in [1,\infty )$, $0\le e^{-x^2}\le e^{-x}$ as $x>1⟹x^2\ge x$ as $e^x$ is increasing
4. Then, consider ${\int }_0^{\infty }{e}^{-x}dx$
5. $={\lim }_A\to \infty {\int }_A^0{e}^{-x}dx={\lim }_{A\to \infty }-e^{-A}+e^{-1}=\frac{1}{e}$
6. This implies that ${\int }_{I}f(x)dx$ converges by Comparison Theorem for Integrals