Chapter 8.4, Problem 34E

To determine

The improper integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ converges or diverges.

Answer to Problem 34E

The integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ converges.

Explanation of Solution

Given information: The improper integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ is given.

Concept used:

1. If $f\left(x\right)$ is continuous on the interval $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ .
2. ${\displaystyle \underset{-a}{\overset{a}{\int }}f\left(x\right)}dx=\left\{\begin{array}{cc}2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)}dx& \text{if }f\left(-x\right)=f\left(x\right)\left(\text{even function}\right)\\ 0& \text{if }f\left(-x\right)=-f\left(x\right)\left(\text{odd function}\right)\end{array}\right.$
3. If $f\left(x\right)$ is continuous, then
${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx+{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx\left(a<c<b\right)$

4. For any two continuous functions $f\left(x\right)$ and $g\left(x\right)$ such that $g\left(x\right)\ge f\left(x\right)\ge 0$ on the interval $\left[a,\infty \right)$ for all $x\ge a$, then

• The improper integral ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx$ converges if ${\displaystyle \underset{a}{\overset{\infty }{\int }}g\left(x\right)}dx$ converges.
• The improper integral ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx$ diverges if ${\displaystyle \underset{a}{\overset{\infty }{\int }}g\left(x\right)}dx$ diverges.

Calculation:

Rewrite and simplify the integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ as shown below:

  $\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}=2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}\\ =2{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{x^4+1}}}+2{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}\end{array}$

The first term on the right side of the obtained equation is a proper integral, while the second term is an improper integral.

Check the convergence of the second term, that is, $2{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ .

Consider

  $f\left(x\right)=2{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ and $g\left(x\right)={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^2}}$ .

Now, determine $g\left(x\right)={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^2}}$ using ${\displaystyle \underset{a}{\overset{\infty }{\int }}g\left(x\right)}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{a}{\overset{b}{\int }}g\left(x\right)}dx$

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^2}}=\underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^2}}dx\\ =\underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{\infty }{\int }}{x}^{-2}}dx\\ =\underset{b\to \infty }{\lim }{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^{\infty }\\ =\underset{b\to \infty }{\lim }{\left[\frac{x^{-2}}{-1}\right]}_1^{\infty }\\ =\underset{b\to \infty }{\lim }{\left[\frac{-1}{x^2}\right]}_1^{\infty }\\ =\underset{b\to \infty }{\lim }\left[-\left(\frac{1}{\infty }-\frac{1}{1}\right)\right]\\ =\underset{b\to \infty }{\lim }\left[-\left(-1\right)\right]\\ =1\end{array}$

This implies that,

Thus, $g\left(x\right)={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^2}}$ converges to $1$ (a finite number).

Since ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^3}}$ converges, it needs to be larger than $2{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ . This implies that $0\le \frac{1}{\sqrt{x^4+1}}\le \frac{1}{x^2}$ on $\left[1,\infty \right)$ .Therefore, $2{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ converges as ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{dx}{x^3}}$ converges.

As a result, the improper integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{dx}{\sqrt{x^4+1}}}$ converges.