mravinjakag

2022-06-21

Antiderivative of $\mathrm{sin}(\frac{1}{x})$

I have to find the values of parameter a so the function

$f:\mathbb{R}\to \mathbb{R},f(x)=\{\begin{array}{ll}\mathrm{sin}\frac{1}{x}& \text{for}x\in \mathbb{R}\setminus \{0\}\\ a& \text{for}x=0.\end{array}$

has an antiderivative on $\mathbb{R}$. I know that $\underset{x\to 0}{lim}\mathrm{sin}\frac{1}{x}$ does not exist in $x=0$. I have found that the answer is $a=0$ but i can't explain it.

I have to find the values of parameter a so the function

$f:\mathbb{R}\to \mathbb{R},f(x)=\{\begin{array}{ll}\mathrm{sin}\frac{1}{x}& \text{for}x\in \mathbb{R}\setminus \{0\}\\ a& \text{for}x=0.\end{array}$

has an antiderivative on $\mathbb{R}$. I know that $\underset{x\to 0}{lim}\mathrm{sin}\frac{1}{x}$ does not exist in $x=0$. I have found that the answer is $a=0$ but i can't explain it.

Jerome Page

Beginner2022-06-22Added 16 answers

Step 1

Suppose F is a primitive of f. Since f is continuous on $\mathbb{R}\setminus \{0\}$, we have

$F(y)-F(x)={\int}_{x}^{y}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$

for $0<x<y$ or $x<y<0$ by the fundamental theorem of calculus. The continuity of F then forces

$F(x)=F(0)+{\int}_{0}^{x}\mathrm{sin}\frac{1}{t}\phantom{\rule{thinmathspace}{0ex}}dt$

for all $x\in \mathbb{R}$. Thus it remains to see that F is differentiable at 0 with derivative 0. Now we have

$\frac{d}{dt}{\textstyle (}{t}^{2}\mathrm{cos}({t}^{-1}){\textstyle )}=\mathrm{sin}({t}^{-1})+2t\mathrm{cos}({t}^{-1})$

and so ${\int}_{0}^{x}\mathrm{sin}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt={\int}_{0}^{x}\frac{d}{dt}{\textstyle (}{t}^{2}\mathrm{cos}({t}^{-1}){\textstyle )}-2t\mathrm{cos}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt={x}^{2}\mathrm{cos}({x}^{-1})-{\int}_{0}^{x}2t\mathrm{cos}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt.$

Step 2

The first part ${x}^{2}\mathrm{cos}({x}^{-1})$ is easily seen to be differentiable at 0 with derivative 0 there, and since

$g(t)=\{\begin{array}{ll}0& ,t=0\\ 2t\mathrm{cos}({t}^{-1})& ,t\ne 0\end{array}$

is continuous on all of $\mathbb{R}$, the fundamental theorem of calculus asserts its differentiability on $\mathbb{R}$, and the derivative is g. Hence ${F}^{\prime}(0)=0-g(0)=0.$

Suppose F is a primitive of f. Since f is continuous on $\mathbb{R}\setminus \{0\}$, we have

$F(y)-F(x)={\int}_{x}^{y}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$

for $0<x<y$ or $x<y<0$ by the fundamental theorem of calculus. The continuity of F then forces

$F(x)=F(0)+{\int}_{0}^{x}\mathrm{sin}\frac{1}{t}\phantom{\rule{thinmathspace}{0ex}}dt$

for all $x\in \mathbb{R}$. Thus it remains to see that F is differentiable at 0 with derivative 0. Now we have

$\frac{d}{dt}{\textstyle (}{t}^{2}\mathrm{cos}({t}^{-1}){\textstyle )}=\mathrm{sin}({t}^{-1})+2t\mathrm{cos}({t}^{-1})$

and so ${\int}_{0}^{x}\mathrm{sin}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt={\int}_{0}^{x}\frac{d}{dt}{\textstyle (}{t}^{2}\mathrm{cos}({t}^{-1}){\textstyle )}-2t\mathrm{cos}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt={x}^{2}\mathrm{cos}({x}^{-1})-{\int}_{0}^{x}2t\mathrm{cos}({t}^{-1})\phantom{\rule{thinmathspace}{0ex}}dt.$

Step 2

The first part ${x}^{2}\mathrm{cos}({x}^{-1})$ is easily seen to be differentiable at 0 with derivative 0 there, and since

$g(t)=\{\begin{array}{ll}0& ,t=0\\ 2t\mathrm{cos}({t}^{-1})& ,t\ne 0\end{array}$

is continuous on all of $\mathbb{R}$, the fundamental theorem of calculus asserts its differentiability on $\mathbb{R}$, and the derivative is g. Hence ${F}^{\prime}(0)=0-g(0)=0.$