1. ## Integral Convergence Theorem

I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?

But I'm still not sure how we can solve this.

Any help would be much appreciated!

2. Originally Posted by kingwinner

I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?

But I'm still not sure how we can solve this.

Any help would be much appreciated!
How rigorous are you trying to be? Letting $z=nx\implies dz=ndx\implies \frac{dz}{n}$. So, our integral becomes $\frac{1}{n}\int_0^{\pi n}\frac{\sin(z)}{z}dz$ and thus $\lim_{n\to\infty}\frac{1}{n}\int_0^{\pi n}\frac{\sin(z)}{z}dz=\lim_{n\to\infty}\frac{1}{n} \cdot\lim_{n\to\infty}\int_0^{\pi n}\frac{\sin(z)}{z}dz=0\cdot\frac{\pi}{2}=0$

If you don't know or can't use the fact that $\int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}$ you can use the following technique.

Let $f(n)=\int_0^{\pi n}\frac{\sin(x)}{x}dx$. Then, $f(n+1)-f(n)=\int_{\pi n}^{\pi(n+1)}\frac{\sin(x)}{x}dx$ but on $[\pi,(n+1)\pi],\text{ }n\in\mathbb{N}$ the sine function is negative and $\frac{1}{x}$ positive and thus $f(n+1)\leqslant f(n)$.

It follows that $\lim_{n\to\infty}\frac{1}{n}\int_0^{\pi n}\frac{\sin(x)}{x}dx\leqslant\lim_{n\to\infty}\fr ac{1}{n}\int_0^{\pi}\frac{\sin(x)}{x}dx=0$ and so the conclusion follows if you can prove...

3. Thanks.

If we instead break the integral into two parts [0,ε], [ε,pi], can we interchange the lim and integral signs to evaluate?

4. Originally Posted by kingwinner
Thanks.

If we instead break the integral into two parts [0,ε], [ε,pi], can we interchange the lim and integral signs to evaluate?
That above theorem, isn't $a\in[a,b]$

Also, do you know the DCT?

5. Let me see if I know enough to solve this...

So if you do want to integrate this by DCT:

So I know that $| \frac { \sin nx }{ nx } | \leq \frac {1}{nx}$ for $x \in (0, \pi )$

Therefore we can place limit inside the integral, so it is:

$\int ^ \pi _0 \lim _{n \rightarrow \infty } \frac { \sin nx }{ nx } dx$ Is this right so far?

Well, then, doesn't it converge to 0?

6. Originally Posted by Drexel28
Let $f(n)=\int_0^{\pi n}\frac{\sin(x)}{x}dx$. Then, $f(n+1)-f(n)=\int_{\pi n}^{\pi(n+1)}\frac{\sin(x)}{x}dx$ but on $[n\pi,(n+1)\pi],\text{ }n\in\mathbb{N}$ the sine function is negative only for odd n and $\frac{1}{x}$ positive and thus $f(n+1)\leqslant f(n)$.
There is a mistake here. However, the first proof was fine (by the way, it doesn't require the value of the integral $\int_0^\infty\frac{\sin x}{x}dx$, just its improper convergence)

Originally Posted by kingwinner
I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?
You can indeed procede this way. For $x>\varepsilon$, for all $n\geq 1$, $\frac{|\sin (nx)|}{nx}\leq \frac{1}{n\varepsilon}$, which implies that the sequence $f_n:x\mapsto \frac{\sin (nx)}{nx}$ converges uniformly to 0 on $[\varepsilon,\pi]$ (but not on $(0,\pi]$ since the limit at 0 is 1 hence the limit is discontinuous). By your theorem, the integral on $[\varepsilon,\pi]$ converges to 0. As for the piece on $[0,\varepsilon]$, you can prove its absolute value is less than $\varepsilon$.

Let me see if I know enough to solve this...

So if you do want to integrate this by DCT:

So I know that $| \frac { \sin nx }{ nx } | \leq \frac {1}{nx}$ for $x \in (0, \pi )$
You should read the statement of the dominated convergence theorem again: you need to dominate the sequence of functions by a function that does not depend on $n$ anymore and that is integrable. Neither of these constraints is fulfilled by your bound. What would work is simply use $| \sin (nx)|\leq nx$ to prove $\left|\frac{\sin (nx)}{nx}\right|\leq 1$, since $1$ is integrable on $[0,\pi]$. This proof using the TCD is much shorter, but much less elementary than the other ones.

7. Originally Posted by Laurent
You can indeed procede this way. For $x>\varepsilon$, for all $n\geq 1$, $\frac{|\sin (nx)|}{nx}\leq \frac{1}{n\varepsilon}$, which implies that the sequence $f_n:x\mapsto \frac{\sin (nx)}{nx}$ converges uniformly to 0 on $[\varepsilon,\pi]$ (but not on $(0,\pi]$ since the limit at 0 is 1 hence the limit is discontinuous). By your theorem, the integral on $[\varepsilon,\pi]$ converges to 0. As for the piece on $[0,\varepsilon]$, you can prove its absolute value is less than $\varepsilon$.
Hi Laurent,

"...but not on $(0,\pi]$ since the limit at 0 is 1 hence the limit is discontinuous..."
Why at x=0, the limit as n->infinity is 1? I think sin(nx)/(nx) isn't even defined when x=0? Can you explain in more detail why it is NOT uniformly convergent on (0,pi], please?

Also, I don't understand how to handle the [0,ε] piece. Can you explain a little more?

8. Originally Posted by kingwinner
Hi Laurent,

"...but not on $(0,\pi]$ since the limit at 0 is 1 hence the limit is discontinuous..."
Why at x=0, the limit as n->infinity is 1? I think sin(nx)/(nx) isn't even defined when x=0? Can you explain in more detail why it is NOT uniformly convergent on (0,pi], please?
Ok, I assumed you let the function equal 1 at 0 since this is the limit (remember $\lim_{x\to0}\frac{\sin x}{x}=1$), but you don't have to.

Suppose the convergence were uniform on $(0,\pi]$. For every $n$, you have $\lim_{x\to 0^+} f_n(x)=1$ like I said. However, you can prove that this, together with the uniform convergence, implies $\lim_{x\to 0^+}f(x)=1$, where $f$ is the limit of $f_n$ (write $|f(x)-1|\leq |f_n(x)-f(x)|+|f_n(x)-1|<2\epsilon$ for suitable choices of large $n$ and small $x$...). In the present case, $f(x)=0$ for $0, hence clearly the previous limit is wrong: this is a contradiction.

Also, I don't understand how to handle the [0,ε] piece. Can you explain a little more?
Just use $\left|\frac{\sin nx}{nx}\right|\leq 1$ inside the integral.

9. Originally Posted by Laurent
Ok, I assumed you let the function equal 1 at 0 since this is the limit (remember $\lim_{x\to0}\frac{\sin x}{x}=1$), but you don't have to.

Suppose the convergence were uniform on $(0,\pi]$. For every $n$, you have $\lim_{x\to 0^+} f_n(x)=1$ like I said. However, you can prove that this, together with the uniform convergence, implies $\lim_{x\to 0^+}f(x)=1$, where $f$ is the limit of $f_n$ (write $|f(x)-1|\leq |f_n(x)-f(x)|+|f_n(x)-1|<2\epsilon$ for suitable choices of large $n$ and small $x$...). In the present case, $f(x)=0$ for $0, hence clearly the previous limit is wrong: this is a contradiction.
OK, but the definition given in my textbook is this...
To show f_n(x) does not converge uniformly to 0 on (0,pi], we need to show that
sup| $f_n(x)-0$| does NOT converge to 0 as n->infinity.
0<x≤pi
How can we prove using this definition?

Just use $\left|\frac{\sin nx}{nx}\right|\leq 1$ inside the integral.
Is this always ≤1?
|sin(nx)/(nx)| 1/nx What if x is very small?

Also, I don't see how this is going to help us to evaluate
ε
∫ sin(nx)/(nx) dx
0
What is the connection?

I hope someone can explain this. Thank you!

10. Originally Posted by kingwinner
OK, but the definition given in my textbook is this...
To show f_n(x) does not converge uniformly to 0 on (0,pi], we need to show that
sup| $f_n(x)-0$| does NOT converge to 0 as n->infinity.
0<x≤pi
How can we prove using this definition?
Here, we have $\sup |f_n(x)-0|\geq 1$ because 1 is the limit when $x\to 0^+$, hence this doesn't converge to 0.

Is this always ≤1?
yes, $|\sin x|\leq x$ for all $x\geq 0$. For instance, because $|\sin x|=\left|\int_0^x -\cos t dt\right|\leq \int_0^x |\cos t|dt\leq \int_0^x dt=x$.

Also, I don't see how this is going to help us to evaluate
ε
∫ sin(nx)/(nx) dx
0
What is the connection?
The point is not to evaluate the integral (you can't) but to prove it is small when $\epsilon$ is small.

The whole scheme is like: let $\epsilon>0$. Using the theorem, we have $\int_\epsilon^\pi f_n(x)dx\to_n 0$ hence for $n\geq n_0$, $\left|\int_\epsilon^\pi f_n(x)dx\right|\leq \epsilon$. Furthermore, $\left|\int_0^\epsilon f_n(x)dx\right|\leq \epsilon$. Thus, $\left|\int_0^\pi f_n(x)dx\right|\leq 2\epsilon$.

11. Originally Posted by Laurent
Here, we have $\sup |f_n(x)-0|\geq 1$ because 1 is the limit when $x\to 0^+$, hence this doesn't converge to 0.

yes, $|\sin x|\leq x$ for all $x\geq 0$. For instance, because $|\sin x|=\left|\int_0^x -\cos t dt\right|\leq \int_0^x |\cos t|dt\leq \int_0^x dt=x$.

The point is not to evaluate the integral (you can't) but to prove it is small when $\epsilon$ is small.

The whole scheme is like: let $\epsilon>0$. Using the theorem, we have $\int_\epsilon^\pi f_n(x)dx\to_n 0$ hence for $n\geq n_0$, $\left|\int_\epsilon^\pi f_n(x)dx\right|\leq \epsilon$. Furthermore, $\left|\int_0^\epsilon f_n(x)dx\right|\leq \epsilon$. Thus, $\left|\int_0^\pi f_n(x)dx\right|\leq 2\epsilon$.

So we showed that
lim ∫ sin(nx)/(nx) dx ε + 0 = ε
n->oo
where ε is some FIXED number between 0 and pi.
note: ∫ is from 0 to pi

But this only gives an upper bound on our answer. How can we find the exact value of lim ∫ sin(nx)/(nx) dx?

Thanks!

12. Originally Posted by kingwinner
So we showed that
lim ∫ sin(nx)/(nx) dx ε + 0 = ε
n->oo
where ε is some FIXED number between 0 and pi.
note: ∫ is from 0 to pi

But this only gives an upper bound on our answer. How can we find the exact value of lim ∫ sin(nx)/(nx) dx?

Thanks!
$\varepsilon>0$ was fixed but arbitrary. So, suppose the integral was non-zero and derive a contradiction.