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.
 A: 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 $t$.
