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Proof of the equation

$$\lim\limits_{A\to \infty } \int_{-A}^{A} \frac{\sin (\alpha x)}{x} f(x) dx = \pi f(0)$$ Preface

This is solely for context

A distribution is a linear operator (a functional) that when acting on a good function (called a test function in the mathematical literature) produces a number. A simple example of a functional is a function F(x) that acts on a good function f(x) as follows

A distribution (or generalised function) D is an equivalent class of weakly convergent sequences of functions [$\psi_n$] and we write

$\int d x D(x) f(x)=\lim _{n \rightarrow \infty} \int_{-\infty}^{\infty} d x f(x) \varphi_{n}(x)$

Two weakly convergent sequences [$\psi_n$] and [$\phi_n$] are equivalent if their difference converges weakly to zero.

$\lim _{\alpha \rightarrow \infty} \frac{\sin (\alpha x)}{\pi x}=\delta(x)$ could someone explain this bit also

in the course textbook the given proof goes as such

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$\begin{aligned} \int_{-A}^{A} d x \frac{\sin (\alpha x)}{x} f(x)=& \int_{-A}^{A} d x \sin (\alpha x) x\left[\frac{f(x)-f(0)}{x}\right] &+f(0) \int_{-A}^{A} d x \frac{\sin (\alpha x)}{x} \end{aligned}$ $\begin{aligned} \int_{-A}^{A} d x \sin (\alpha x) \psi(x)=&-\left.\frac{\cos (\alpha x)}{\alpha} \psi(x)\right|_{-A} ^{A} &+\frac{1}{\alpha} \int_{-A}^{A} d x \cos (\alpha x) \frac{d \psi(x)}{d x}, \end{aligned}$

Now we define

$\\ \psi(x)=(f(x)-f(0)) / x \\$

which is infinitely differentiable and its value and the value of its derivatives at zero can be calculated by L’Hospital rule. Then, integrating by parts the first integral we have.

I dont understand why the derivative at 0 is relevant although $\frac{\sin (\alpha x)}{x}$ attains a maximum at 0 finally the integration works out fine if $\alpha=1$ and $f(x)=1$ and yields the answer $\pi$.

proof for above statement

sinx/x is even so we multiply the integral by a factor of 2.

EDIT

I now have a mediocre understanding of delta function but im still unable to figure out why I cant compute the integral $\quad\quad\quad\quad\quad\quad\quad\quad\quad\int_{-\infty}^{\infty}e^{-x^2}(x^2+1)dx$

the answer is $\frac{3}{2}\sqrt{\pi}$ but if we substitute f(0)=1 where f(x)=x^2+1 and then just integrate the gamma function we get just $\sqrt{\pi}$

Here is the link to the paper

link to the paper

area of concern

The Schwartz Definition 3.1

Any advice would be great.

Thanks once again.

Monocerotis
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  • What is the hypothesis on $f$? – mathcounterexamples.net May 14 '22 at 08:04
  • I have edited the question – Monocerotis May 14 '22 at 08:36
  • @mathcounterexamples.net is there anyway I can improve my question.Would you then perhaps consider answering it? – Monocerotis May 18 '22 at 06:36
  • If I were to guess, you have a typo and it should instead be $$\lim\limits_{A\to \infty } \int_{-A}^{A} \frac{\sin (A x)}{x} f(x) dx = \pi f(0)$$ where $f$ is an appropriate test function (in this case, I don't know how general you can make it, but $C^\infty$ with compact support seems to suffice) – Brian Moehring May 18 '22 at 07:45
  • No that's not what it says. Let me link the paper I was reading. – Monocerotis May 18 '22 at 07:58
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    While not in the same exact form given in the paper, the limit I gave is equivalent to the one in the paper. Again, you have a typo. – Brian Moehring May 18 '22 at 08:10
  • does this particular identity hold true for all $\alpha$ or only just for $ \lim _{\alpha \rightarrow \infty} $ because it surely holds for 1 and by the property of the delta function $\delta(\alpha x)=1/\alpha(\delta(x)$ it must hold true for all real numbers – Monocerotis May 18 '22 at 08:18
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    It doesn't necessarily hold true for $\alpha = 1$. It really only holds as $\alpha \to \infty$. – Brian Moehring May 18 '22 at 08:19
  • The paper has a typo and you have misinterpreted. It is the limit as $\alpha\to\infty$ as Brian says – FShrike May 18 '22 at 11:58
  • And technically they take $\lim_{A\to\infty}\lim_{\alpha\to\infty}$, whether or not you can interchange the limits I don't know – FShrike May 18 '22 at 11:59
  • @FShrike Technically in the paper, $A$ is a chosen constant depending on the support of $f$, but if we were to re-interpret it as a value we're taking the limit of, the order for the limits would be $\lim_{\alpha\to\infty}\lim_{A\to\infty}$. We can interchange the limits, but it's unnecessary for the given argument. – Brian Moehring May 18 '22 at 18:07
  • @BrianMoehring They take $\alpha\to\infty$ first no? – FShrike May 18 '22 at 18:08
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    @FShrike No. First, they choose $A$ such that the support of $f$ is contained in $[-A,A]$. The idea is to choose $A$ such that $\int_{-\infty}^\infty [\ldots] = \int_{-A}^A [\ldots]$. To make it this choose look more like a limit, you could think of this as choosing an $A$ such that $\int_{-\infty}^\infty [\ldots] = \int_{-a}^a [\ldots]$ for all $a \geq A$. – Brian Moehring May 18 '22 at 18:10
  • @Monocerotis This may sound harsh, but after reading your post several times, I have to say: What is your actual question? You seem to have a lot of misunderstandings, from what the actual limit is, to what is being shown, to even how to read a two-column paper (aside from one paragraph that is mostly unrelated to the post, all of it comes from "3.1.2 The Temple Definition", not the Schwartz definition, which it seems you just continuted from the first column of the previous page). Taken together, at least for me, it becomes almost impossible to discern what precisely you're asking about. – Brian Moehring May 18 '22 at 19:39
  • I'm so sorry about all the trouble.I would like to clarify.We take a convergent integral and multiply it by a scaling factor gamma which infinitely increases the functions height at it's maxima and we also multiply the argument of the function with gamma which compresses the function laterally so f(0) is 0 everywhere except 0 and this is our delta function (distribution ) – Monocerotis May 19 '22 at 02:51
  • Also since the scaling factor implies that f is 0 at all points except 1 the limits of the integration could be anything so long as it has compact support.Thanks a lot for your input.I learned a lot and am still learning – Monocerotis May 19 '22 at 02:58
  • Are you sure that's the result you're looking for? Cause if you apply complex analysis the result should be $ \pi i f(0) sin(0) $ – Oleatthewheel Aug 27 '22 at 04:32

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