Caitlyn Cole

2022-05-02

The antiderivative of sin(1/x)

How to prove that the function $f(x)=\mathrm{sin}\frac{1}{x}$ for $x\ne 0,f(0)=0$ has an antiderivative? This means $F(x)={\int}_{0}^{x}\mathrm{sin}(1/t)dt$ has derivative 0 at $x=0$, but I have no idea how to prove it.

How to prove that the function $f(x)=\mathrm{sin}\frac{1}{x}$ for $x\ne 0,f(0)=0$ has an antiderivative? This means $F(x)={\int}_{0}^{x}\mathrm{sin}(1/t)dt$ has derivative 0 at $x=0$, but I have no idea how to prove it.

hadnya1qd

Beginner2022-05-03Added 15 answers

Step 1

I just blogged another solution to this question in my blog. The idea is to consider the following function

$G(x)=\{\begin{array}{ll}{x}^{2}\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

which is differentiable. Clearly,

${G}^{\prime}(x)=\{\begin{array}{ll}\mathrm{sin}\frac{1}{x}+2x\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

Step 2

Hence, ${G}^{\prime}=f+h$ where

$h(x)=\{\begin{array}{ll}2x\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

Since h is continuous, it has antiderivative H, thus giving us $f=(G-H{)}^{\prime}$. In other words, $G-H$ is an antiderivative of f.

I just blogged another solution to this question in my blog. The idea is to consider the following function

$G(x)=\{\begin{array}{ll}{x}^{2}\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

which is differentiable. Clearly,

${G}^{\prime}(x)=\{\begin{array}{ll}\mathrm{sin}\frac{1}{x}+2x\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

Step 2

Hence, ${G}^{\prime}=f+h$ where

$h(x)=\{\begin{array}{ll}2x\mathrm{cos}\frac{1}{x},& \text{if}x\ne 0,\\ 0,& \text{if}x=0.\end{array}$

Since h is continuous, it has antiderivative H, thus giving us $f=(G-H{)}^{\prime}$. In other words, $G-H$ is an antiderivative of f.

wellnesshaus4n4

Beginner2022-05-04Added 24 answers

Explanation:

We can substitute $u={t}^{-1}$ to get a more convenient expression:

$\begin{array}{rl}\left|\frac{F(x)-F(0)}{x}\right|& =\frac{1}{|x|}|{\int}_{0}^{x}\mathrm{sin}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt|\\ \text{(symmetry)}& & =\frac{1}{|x|}\left|{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{sin}u}{{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du\right|& =\frac{1}{|x|}|{[-\frac{\mathrm{cos}u}{{u}^{2}}]}_{1/|x|}^{\mathrm{\infty}}-2{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{cos}u}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du|\\ & =\frac{1}{|x|}||x{|}^{2}\mathrm{cos}\frac{1}{|x|}-2{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{cos}u}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du|\\ & \u2a7d|x|+\frac{1}{|x|}{\int}_{1/|x|}^{\mathrm{\infty}}\frac{2}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du\\ & =2|x|.\end{array}$

Thus ${F}^{\prime}(0)=0$

We can substitute $u={t}^{-1}$ to get a more convenient expression:

$\begin{array}{rl}\left|\frac{F(x)-F(0)}{x}\right|& =\frac{1}{|x|}|{\int}_{0}^{x}\mathrm{sin}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt|\\ \text{(symmetry)}& & =\frac{1}{|x|}\left|{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{sin}u}{{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du\right|& =\frac{1}{|x|}|{[-\frac{\mathrm{cos}u}{{u}^{2}}]}_{1/|x|}^{\mathrm{\infty}}-2{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{cos}u}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du|\\ & =\frac{1}{|x|}||x{|}^{2}\mathrm{cos}\frac{1}{|x|}-2{\int}_{1/|x|}^{\mathrm{\infty}}\frac{\mathrm{cos}u}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du|\\ & \u2a7d|x|+\frac{1}{|x|}{\int}_{1/|x|}^{\mathrm{\infty}}\frac{2}{{u}^{3}}\phantom{\rule{thinmathspace}{0ex}}du\\ & =2|x|.\end{array}$

Thus ${F}^{\prime}(0)=0$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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