mravinjakag

2022-06-21

Antiderivative of $\mathrm{sin}\left(\frac{1}{x}\right)$
I have to find the values of parameter a so the function

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

Step 1
Suppose F is a primitive of f. Since f is continuous on $\mathbb{R}\setminus \left\{0\right\}$, we have
$F\left(y\right)-F\left(x\right)={\int }_{x}^{y}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
for $0 or $x by the fundamental theorem of calculus. The continuity of F then forces
$F\left(x\right)=F\left(0\right)+{\int }_{0}^{x}\mathrm{sin}\frac{1}{t}\phantom{\rule{thinmathspace}{0ex}}dt$
for all $x\in \mathbb{R}$. Thus it remains to see that F is differentiable at 0 with derivative 0. Now we have
$\frac{d}{dt}\left({t}^{2}\mathrm{cos}\left({t}^{-1}\right)\right)=\mathrm{sin}\left({t}^{-1}\right)+2t\mathrm{cos}\left({t}^{-1}\right)$
and so ${\int }_{0}^{x}\mathrm{sin}\left({t}^{-1}\right)\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{x}\frac{d}{dt}\left({t}^{2}\mathrm{cos}\left({t}^{-1}\right)\right)-2t\mathrm{cos}\left({t}^{-1}\right)\phantom{\rule{thinmathspace}{0ex}}dt={x}^{2}\mathrm{cos}\left({x}^{-1}\right)-{\int }_{0}^{x}2t\mathrm{cos}\left({t}^{-1}\right)\phantom{\rule{thinmathspace}{0ex}}dt.$
Step 2
The first part ${x}^{2}\mathrm{cos}\left({x}^{-1}\right)$ is easily seen to be differentiable at 0 with derivative 0 there, and since
$g\left(t\right)=\left\{\begin{array}{ll}0& ,t=0\\ 2t\mathrm{cos}\left({t}^{-1}\right)& ,t\ne 0\end{array}$
is continuous on all of $\mathbb{R}$, the fundamental theorem of calculus asserts its differentiability on $\mathbb{R}$, and the derivative is g. Hence ${F}^{\prime }\left(0\right)=0-g\left(0\right)=0.$

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