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How to solve $\left(\int_0^x t\sin(\frac{1}{t})dt\right)’_{x=0}$?

I’m not sure if I can use Newton-Leibniz Theorem $$\dfrac{d}{dx}\left(\int_a^xf(t) dt=f(x)\right)$$ because $f(x)=x\sin(\frac{1}{x})$ is undefined where $x=0$. Is it correct that let $f(0)=0$?

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    $\begingroup$ What are the conditions required to apply the theorem ? $\endgroup$
    – user65203
    Nov 20, 2017 at 9:21
  • $\begingroup$ The problem becomes more interesting if $t\sin(1/t)$ is replaced by $\sin(1/t)$ or $\cos(1/t)$. In this case the discontinuity at $0$ can not be removed and it requires some more manipulation to get the answer as $0$. $\endgroup$
    – Paramanand Singh
    Nov 21, 2017 at 3:34

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The fundamental theorem of calculus only requires your function to be continuous over a closed domain ; in our case $f : t \mapsto t \sin(1 / t)$ is continuous on $(0, a]$ for any $a$, so we can instead use the continuous continuation $\widetilde{f}$ which is equal to $f$ over $\mathbb{R}^*_+$ and is $0$ for $x = 0$ (this is indeed the limit of $f$, as $\sin$ is bounded by $1$). It turns out that $f$ and $\widetilde{f}$ have the same integral over any closed interval of $\mathbb{R}^*_+$ ; can you take it from there ?

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  • $\begingroup$ I see. But use the fundamental theorem I get the $F’(x)=x\sin{\dfrac{1}{x}}$. To get $F’(0)$, can I let $x\to 0$, directly? $\endgroup$
    – 闫嘉琦
    Nov 20, 2017 at 9:43
  • $\begingroup$ Absolutely, you can just do that. $\endgroup$ Nov 20, 2017 at 9:46
  • $\begingroup$ Because $f$ is continuous? Thanks a lot. $\endgroup$
    – 闫嘉琦
    Nov 20, 2017 at 9:47
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    $\begingroup$ As I said, $f$ is not continuous over a closed interval because it is not defined at $0$. Instead, we consider the continuation, which is in turn defined at $0$ and stays continuous. Since this continuation is continuous and only differs from $f$ at one single point, both integrals are equal and you can apply the fundamental theorem of calculus to the continuation, leading to your answer. $\endgroup$ Nov 20, 2017 at 9:49
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Hint. Let $F(x):=\int_0^x t\sin(1/t)dt$, then $F(0)=0$ (note that $\lim_{t\to 0^+} t\sin(1/t)=0$ which means that the integrand function is bounded in a neighbourhood of $0$). Hence $F$ is a continuous function and $$\left(\int_0^x t\sin(1/t)dt\right)’_{x=0}=\lim_{x\to 0}\frac{F(x)-F(0)}{x-0}=\lim_{x\to 0}\frac{1}{x}\int_0^x t\sin(1/t)dt.$$ Now we may use L'Hopital rule (and Newton-Leibniz Theorem for $x\not=0$). You may also use the Mean Value Theorem: for any $x\not=0$ there is $t_x$ between $0$ and $x$ such that $$\frac{1}{x}\int_0^x t\sin(1/t)=t_x\sin(1/t_x).$$

P.S. More generally if a derivative of a continuous function has a limit, it must agree with that limit. See prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$.

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  • $\begingroup$ Mh, how can the limit be evaluated ? Certainly not by L'Hospital... $\endgroup$
    – user65203
    Nov 20, 2017 at 9:25
  • $\begingroup$ This is wrong. $t \sin(1 / t)\underset{t \rightarrow 0}{\rightarrow} 0$, because $sin$ is bounded by 1 for any t. $\endgroup$ Nov 20, 2017 at 9:27
  • $\begingroup$ @Matrefeytontias Thanks, I corrected the typo. $\endgroup$
    – Robert Z
    Nov 20, 2017 at 9:33
  • $\begingroup$ Use Lhospital rule then it follows $F’(0)=(\int_0^x t\sin{\dfrac{1}{t}})’$, it is just the question itself. $\endgroup$
    – 闫嘉琦
    Nov 20, 2017 at 9:37
  • $\begingroup$ No, by Hopital we evaluate $\lim_{x\to 0}F'(x)$! $\endgroup$
    – Robert Z
    Nov 20, 2017 at 9:40
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Let , $$f(x) = \int_0^x t\sin(1/t)dt$$

$$\left(\int_0^x t\sin(1/t)dt\right)’_{x=0}=f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{1}{x}\int_0^x t\sin(1/t)dt.$$

But since, $|\sin(1/t)|\le 1$

$$\left|\lim_{x\to 0}\frac{1}{x}\int_0^x t\sin(1/t)dt\right|\le \lim_{x\to 0}\frac{1}{x}\int_0^x tdt = \lim_{x\to 0}\frac{x}2 =0$$

Thus, $$f'(0)=\left(\int_0^x t\sin(1/t)dt\right)’_{x=0} = 0$$

Also see here for similar problem: Compute $f'(0)$ and check if $f'$ is continuous or not

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