# Thread: Proof involving an integral of a product of 2 functions

1. ## Proof involving an integral of a product of 2 functions

Let $\displaystyle n\in \mathbb{N}$, consider $\displaystyle \Psi _n (x)=\frac{\sin (nx)}{\pi x}$ for $\displaystyle x\neq 0$. Let $\displaystyle \Psi _n (0)=\frac{n}{\pi}$ such that $\displaystyle \Psi _n$ is continuous.
Demonstrate that $\displaystyle \lim _{n\to \infty} \int _{\mathbb{R}} \Psi _n (x) f(x)dx=f(0)$ for any functions $\displaystyle f$ that are infinitely many times differentiable and with compact support.
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Believe it or not, this exercise is a problem I've in my electromagnetism course. We have not been introduced what a compact support is so I checked out in wikipedia and it states Originally Posted by wikipedia
the support of a function is the set of points where the function is not zero, or the closure of that set.
which seems 2 different things to me.

Do you have any tip, hint or idea about how to do the proof? I'm willing to put a lot of efforts to do it. I just don't know how to tackle it.

2. Originally Posted by arbolis Let $\displaystyle n\in \mathbb{N}$, consider $\displaystyle \Psi _n (x)=\frac{\sin (nx)}{\pi x}$ for $\displaystyle x\neq 0$. Let $\displaystyle \Psi _n (0)=\frac{n}{\pi}$ such that $\displaystyle \Psi _n$ is continuous.
Demonstrate that $\displaystyle \lim _{n\to \infty} \int _{\mathbb{R}} \Psi _n (x) f(x)dx=f(0)$ for any functions $\displaystyle f$ that are infinitely many times differentiable and with compact support.
First, you can notice that the change of variable $\displaystyle y=nx$, together with letting $\displaystyle n$ tend to infinity inside the integral gives a "formal" proof, provided you know $\displaystyle \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$. Maybe this is enough for your electromagnetic course.

If this were maths, you'd have to justify taking the limit inside the integral. One difficulty is that the limit is an improper integral, hence usual Lebesgue's theorem don't apply directly. Here's a suggestion. On $\displaystyle [a,b]$ containing the support of $\displaystyle f$, write $\displaystyle f(x)=f(0)+x\varphi(x)$ where $\displaystyle \varphi$ is $\displaystyle \mathcal{C}^\infty$ as well (but not necessarily compactly supported). Use linearity to split the integral; for the first part, change of variable, use the above classic integral, ok. For the second part, you can note this is a case of Riemann-Lebesgue lemma. It can be proved for step functions $\displaystyle \varphi$ first and generalized by density, for instance. Or using an integration by part (derivating phi), if I remember well.

3. Originally Posted by Laurent First, you can notice that the change of variable $\displaystyle y=nx$, together with letting $\displaystyle n$ tend to infinity inside the integral gives a "formal" proof, provided you know $\displaystyle \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$. Maybe this is enough for your electromagnetic course.

If this were maths, you'd have to justify taking the limit inside the integral. One difficulty is that the limit is an improper integral, hence usual Lebesgue's theorem don't apply directly. Here's a suggestion. On $\displaystyle [a,b]$ containing the support of $\displaystyle f$, write $\displaystyle f(x)=f(0)+x\varphi(x)$ where $\displaystyle \varphi$ is $\displaystyle \mathcal{C}^\infty$ as well (but not necessarily compactly supported). Use linearity to split the integral; for the first part, change of variable, use the above classic integral, ok. For the second part, you can note this is a case of Riemann-Lebesgue lemma. It can be proved for step functions $\displaystyle \varphi$ first and generalized by density, for instance. Or using an integration by part (derivating phi), if I remember well.
I hope the informal way is enough but I'm not really sure, my professor doesn't consider himself a physicist though he is one but I agree he looks like a mathematical physicist or so.
So putting $\displaystyle y=nx$, $\displaystyle \lim _{n\to \infty}$ converts into $\displaystyle \lim _{y\to \infty}$.
Putting the limit inside the integral, we obtain $\displaystyle \int _{\mathbb{R}} \lim _{y \to \infty} \frac{\sin (y) f(x)}{\pi x} dx$.
Should I replace $\displaystyle x$ by $\displaystyle \frac{y}{n}$ to reach something involving $\displaystyle \frac{\sin y}{y}$?
Also the integral is over $\displaystyle \mathbb{R}$, so $\displaystyle \int _{\mathbb{R}} \frac{\sin x}{x} dx =\pi$. Why did you took only the positive real numbers instead of the whole set?

4. Originally Posted by arbolis I hope the informal way is enough but I'm not really sure, my professor doesn't consider himself a physicist though he is one but I agree he looks like a mathematical physicist or so.
So putting $\displaystyle y=nx$, $\displaystyle \lim _{n\to \infty}$ converts into $\displaystyle \lim _{y\to \infty}$.
Putting the limit inside the integral, we obtain $\displaystyle \int _{\mathbb{R}} \lim _{y \to \infty} \frac{\sin (y) f(x)}{\pi x} dx$.
You shouldn't mix variables inside the integral. Putting $\displaystyle y=nx$, we have $\displaystyle \int \frac{\sin(nx)}{\pi x}f(x)dx= \int \frac{\sin y}{\pi y}f\left(\frac{y}{n}\right)dy$, and we still want $\displaystyle n\to\infty$ (by the way, "y" is just a dummy variable). By continuity at 0, the "f" term converges to $\displaystyle f(0)$.

Also the integral is over $\displaystyle \mathbb{R}$, so $\displaystyle \int _{\mathbb{R}} \frac{\sin x}{x} dx =\pi$. Why did you took only the positive real numbers instead of the whole set?
It was just because this is the way I remember the formula, but sure it is on the whole line here.

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