Start by using integration by parts, then use the fact that since $f'$ is continuous, there is a constant $M$ such that $M \geq |f'(x)| \forall x\in[0,1]$.

I used integration by parts and got:

$$ I=\lim_{n \to \infty} \int_{0}^{1} f(x) \sin(nx)dx=\lim_{n \to \infty} \left[-\dfrac{1}{n} \int_{0}^{1} f'(x) \cos(nx)dx \right] $$

Stuck now. No clue how to use that $M$ information with the squeeze theorem. My instinct would be to assume $M>f'(x)$ such that: $$ \int_{0}^{1} (-M) \cos(nx)dx \leq \int_{0}^{1} f'(x) \cos(nx)dx \leq \int_{0}^{1} (M) \cos(nx)dx $$

Not sure if I bounded the above correctly, but then I'd do:

$$ \left[ -\dfrac{M\sin(nx)}{n} \right]_{x=0}^{x=1} \leq \int_{0}^{1} f'(x) \cos(nx)dx \leq \left[ \dfrac{M\sin(nx)}{n} \right]_{x=0}^{x=1} $$ $$ -\dfrac{M\sin(n)}{n} \leq \int_{0}^{1} f'(x) \cos(nx)dx \leq \dfrac{M\sin(n)}{n} $$

Then I divide the whole thing by $n$, and take the limit as $n \to \infty$ such that the middle part of the inequality becomes $I$.

$$ -\lim_{n\to\infty}\dfrac{M\sin(n)}{n^2} \leq I \leq \lim_{n\to\infty}\dfrac{M\sin(n)}{n^2} $$

Evaluating the limits I get

$$ 0 \leq I \leq 0 $$

Thus, $I=0$. Not sure if this is the right approach.

This question has been answered, but using different methods out of the scope of this class. Can someone please solve it using the above hint.

  • $\begingroup$ What to show !? $\endgroup$
    – Z Ahmed
    Commented Sep 4, 2020 at 5:57
  • $\begingroup$ Please show your effort. $\endgroup$ Commented Sep 4, 2020 at 5:58
  • $\begingroup$ Just updated the post. $\endgroup$
    – user256872
    Commented Sep 4, 2020 at 6:00
  • $\begingroup$ $|\int_0^{1}f'(x)\cos (nx)dx|\leq \int_0^{1} (M)(1)dx=M$. $\endgroup$ Commented Sep 4, 2020 at 6:12
  • $\begingroup$ So from there do I just say that the right side goes to 0 as n goes to infinity, therefore the left side is bounded between 0 and 0? $\endgroup$
    – user256872
    Commented Sep 4, 2020 at 6:17

1 Answer 1


Your attempt is pretty much it. Here is how I would write it down:

Integrating by parts,

$$\int_0^1 f(x) \sin(nx) dx = \frac{-1}{n}\int_0^1 f(x) d \cos(nx)$$ $$= \frac{-1}{n}f(x) \cos(nx)\big\vert_{x=0}^{x=1} + \frac{1}{n} \int_0^1 \cos(nx) f'(x) dx$$

Now, $$\frac{-1}{n}f(x) \cos(nx)\big\vert_{x=0}^{x=1}= \frac{-1}{n} f(1) \cos(n) + \frac{1}{n} f(0) \stackrel{n \to \infty}\longrightarrow 0$$

and $$\left|\frac{1}{n}\int_0^1 \cos(nx) f'(x) dx\right| \leq \frac{1}{n} \int_0^1 |f'(x)|dx \leq \frac{1}{n} \Vert f' \Vert_\infty \stackrel{n \to \infty}\longrightarrow 0$$ where we used that $f'$ is bounded since it is continuous on the compact interval $[0,1]$.

Combining all this, we see that $$\lim_{n\to \infty} \int_0^1 f(x) \sin(nx) dx = 0$$


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