# Finding $f'(0)$ when $f(x)=\int\limits_0^x\sin\left(\frac{1}{t}\right)dt$

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?

-
Do you mean, show that $f'(x)=0$ for some $x$? Because $f'(x)\neq 0$ for most values of $x$ –  Thomas Andrews Jun 27 '12 at 21:17
Sorry, that is my inadvertence –  user34574 Jun 27 '12 at 21:22
Sorry, you are right... again –  user34574 Jun 27 '12 at 21:30
$f'(0)=\lim\limits_{h\to0}\dfrac{f(0+h)-f(0)}{h}$. Probably you'll need to find the limit by "squeezing". –  Michael Hardy Jun 27 '12 at 21:35
How to find this limit? L'Hopital's rule doesn't work too! –  user34574 Jun 27 '12 at 21:38

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 $\left|\dfrac{f(x)}{x}\right|\leq 2|x|$.
-
Then $0\leq |f'(0)|\leq\lim\limits_{x\to 0} 2|x|=0$. Right? –  user34574 Jun 27 '12 at 21:48
I do not entirely follow the logic of your comment, but e.g. you could use the so-called squeeze theorem, observing that $0<|f(x)/x|<2|x|$ and $\lim\limits_{x\to 0}2|x| = 0$ implies that $\lim\limits_{x\to 0}\frac{f(x)}{x}=0$, hence by definition $f'(0)$ exists and is $0$. –  Jonas Meyer Jun 27 '12 at 21:53
Note that if you could assume that $f'(0)$ exists then this would be very easy, because $f$ is an even function so $f'(0)=0$ if it exists. ($f(x)=f(-x)\implies f'(x)=-f'(-x)$.) –  Jonas Meyer Jun 27 '12 at 21:53
You are so smart! –  user34574 Jun 27 '12 at 21:54

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 $x\to 0$.]

-
Your solution is more detailed but Jonas Meyer for the first. –  user34574 Jun 27 '12 at 21:55
@Jataga — Detailed solutions are longer to write ! (Just teasing you, I don't mind your choice ;)) –  Lierre Jun 27 '12 at 22:02
Ok I've got another unanswered question. I will be glad to accept your solution :D –  user34574 Jun 27 '12 at 22:04

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

-