Chapter 5, Problem 12RQ

To determine

To Find: whether the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{\sqrt{2}}}}dx$ is convergent

Answer to Problem 12RQ

The limit exists as a finite number and so the improper integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{\sqrt{2}}}}dx$ is convergent

Explanation of Solution

Given information: we have been given ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{\sqrt{2}}}}dx$

the improper integrals ${\displaystyle \underset{a}{\overset{\infty }{\int }}f(x)dx}$ is called convergent if the corresponding limit exist and

Divergent if the limit does not exist.

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^{\sqrt{2}}}}dx$

Rewrite the improper integral as a limit

  $\underset{t\to \infty }{\lim }{\displaystyle \underset{1}{\overset{t}{\int }}\frac{1}{x^{\sqrt{2}}}dx}$

We can also write it like this

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

Evaluate at $x=1$ and $x=t$

  $\frac{1}{-\sqrt{2}+1}\underset{t\to \infty }{\lim }\left(t^{-\sqrt{2}+1}-1\right)$ [since, $-\sqrt{2}+1<0$ ]

Therefore,

  $\begin{array}{l}\underset{t\to \infty }{\lim }\left(t^{-\sqrt{2}+1}\right)=0\\ \frac{1}{-\sqrt{2}+1}\left(0-1\right)= \frac{1}{\sqrt{2}-1}\end{array}$

The limit exists as a finite number so the integral is convergent