# Proof of limit that involves integral

I was trying to prove the following statement but I am not sure whether my proof is correct or not.

Prove that for any $$a > 0$$ the following holds:

$$\lim_{t \to \infty} \int_t^{t+a}\frac{sin(x)}{x}dx = 0$$

My approach:

Denote $$f(t) = \int_t^{t+a}\frac{sin(x)}{x}dx$$ , Then we need to prove that for any $$\epsilon > 0$$ there is $$M \in R$$ such that if $$t > M$$ then $$|f(t) - 0| < \epsilon$$

Let $$\epsilon = \frac{\epsilon}{a}$$, from the fact that $$\lim_{x \to \infty} \frac{sin(x)}{x} = 0$$ we know that there is an element $$M1 \in R$$ such that if $$x>M1$$ then $$|\frac{sin(x)}{x}| < \frac{\epsilon}{a}$$

Therefore for the same $$M1$$ if $$t>M1$$ then for any $$x \in [t,t+a]$$ :

$$|\frac{sin(x)}{x}| < \frac{\epsilon}{a} \Longleftrightarrow -\frac{\epsilon}{a} < \frac{sin(x)}{x} < \frac{\epsilon}{a}$$

Therefore $$f$$ is bounded between $$-\frac{\epsilon}{a}$$ and $$\frac{\epsilon}{a}$$, so:

$$-\frac{\epsilon}{a} (t+a-t) < \int_t^{t+a}\frac{sin(x)}{x}dx < \frac{\epsilon}{a} *(t+a-t) \Longleftrightarrow -\epsilon < \int_t^{t+a}\frac{sin(x)}{x}dx < \epsilon \Longleftrightarrow$$

$$|\int_t^{t+a}\frac{sin(x)}{x}dx| < \epsilon$$

This is what I did but I am not sure that this is true. I will appreciate any help, thanks in advance.

• Looks good to me. I would fix up the formatting, but good work otherwise! Commented Nov 3, 2021 at 20:54
• @CameronWilliams Thank you :) Commented Nov 3, 2021 at 21:00
• As mentioned in the other comments the proof looks fine to me too. But I think it will be better if you mention , for all finite $a>0$ Commented Nov 4, 2021 at 11:03

You did well so I'm just being picky here.

You wrote

Then we need to prove that for any $$\epsilon > 0$$ there is $$M \in R$$ such that if $$t > M$$ then $$|f(t) - 0| < \epsilon$$

The usual definition of the limit at infinity usually demands $$M> 0$$ and not just $$M \in \mathbb R$$.

At the next line you wrote

Let $$\epsilon = \frac{\epsilon}{a}$$, ...

I'm not a fan of this as it sort of implies that $$a = 1$$. Rather you should start your proof with something like this

Let $$\epsilon > 0$$. In particular $$\epsilon' = \frac{\epsilon}{a}> 0$$ so $$\exists M_1 >0$$ such that for $$x > M_1$$ we have $$\left \vert \frac {\sin(x)} {x}\right\vert< \frac{\epsilon}{a}$$

Later you also wrote

$$|\frac{sin(x)}{x}| < \frac{\epsilon}{a} \Longleftrightarrow -\frac{\epsilon}{a} < \frac{sin(x)}{x} < \frac{\epsilon}{a}$$ Therefore $$f$$ is bounded between $$-\frac{\epsilon}{a}$$ and $$\frac{\epsilon}{a}$$,

To be picky you defined $$f$$ as $$f(t) = \int_t^{t+a} \frac{\sin(x)}{x}dx$$ so in fact it is $$\frac{\sin (x)}{x}$$ that is between $$-\frac{\epsilon}{a}$$ and $$\frac \epsilon a$$ not $$f$$! Notice that you then show that $$f$$ is actually between $$-\epsilon$$ and $$\epsilon$$.

All in all you did pretty well just some minor mistakes :)

I think the exercice would have been more straightforward if you just used the property $$\left\vert \int g(x) dx \right\vert \leq \int \vert g(x) \vert dx.$$ Since you would then have had $$\left\vert \int_t^{t+a} \frac{\sin (x)}{x}dx\right\vert \leq \int_t^{t+a}\left\vert \frac{\sin(x)}{x}\right\vert dx < \int_t^{t+a} \frac{\epsilon}{a}dx = \epsilon$$ for large enough $$t$$.