17
$\begingroup$

I need to show that $f'(0)=0$ for $ f(x)=\int\limits_0^x\sin\left(\frac{1}{t}\right)dt $ But fundamental theorem of calculus is unapplicable here. What should I do?

  • 0
    Related: http://math.stackexchange.com/questions/163717/find-the-derivative-of-the-primitive-of-a-discontinous-function2012-06-28

3 Answers 3

9

Here is one approach:

  • Use integration by parts to show that $f(x)=x^2\cos(1/x)-\int_0^x 2t\cos(1/t)dt$ for $x\neq 0$.
  • Use this to show that $\left|\dfrac{f(x)}{x}\right|\leq 2|x|$.
  • 2
    You are so smart!2012-06-27
7

To strengthen the convergence of an integral, a integration by parts is always a good idea. Here we have $\begin{aligned} f(x) &=\int_0^x\sin\left(\frac 1t \right) dt\\ &= \left[ t^2 \cos\left(\frac 1t \right)\right]_0^x - \int_0^x 2t\cos\left(\frac 1t \right)dt\\ &= x^2 \cos\left(\frac 1x \right) - \int_0^x 2t\cos\left(\frac 1t \right)dt \end{aligned}$ You can apply the fundamental theorem of calculus for the second term, since it is the integral of a continuous function. The first term is a $O(x^2)$, so it is differentiable at zero with null derivative. In the end, we get $f'(0) = 0$.

[Reminder — A function $f$ is derivable at zero with derivative $a$ if and only if $f(x) = f(0) + ax + o(x)$ when $x\to 0$.]

  • 0
    Ok I've got another $u$nanswered q$u$estion. I will be glad to accept your solution :D2012-06-27
0

You can use the fundamental theorem of calculus to show that $f$ is differentiable in an interval around $0$ and apply Rolle's theorem.