Limit in distributions of $\frac{\sin(tx)}{x}$

Question

How can I find the limit of $\frac{\sin(tx)}{x}$ as $t \to \infty$ in $D'$ ? I understand that i need to see the $\lim_{t \to \infty}{\int_{\infty}^{\infty}{\frac{\sin(tx)\phi(x)}{x}dx}}$ for every test function $\phi$ in $D$. Any help would be appreciated.

Answer 1

Using Fourier transform

Clearly, $(\xi \mapsto \frac{\sin t\xi}{\xi}) \in S'(\mathbb R) \subset D'(\mathbb R),$ so we can use Fourier transforms. Now, $\frac{\sin t\xi}{\xi} = \mathcal{F}\{\frac12 \chi_{[-t,t]}(x)\},$ so since $\frac12 \chi_{[-t,t]}(x) \to \frac12$ (constant function) as $t \to \infty$ we have that $\frac{\sin t\xi}{\xi} \to \mathcal{F}\{\frac12\} = \pi\delta(\xi)$ in $S'$ as $t \to \infty.$ But convergence in $S'$ implies convergence in $D'.$ Thus, $x \mapsto \frac{\sin tx}{x} \to \pi\delta(x)$ in $D'.$

Calculation of $\mathcal{F}\{\chi_{[-a,a]}(x)\}$: $$ \mathcal{F}\{\chi_{[-a,a]}(x)\} = \int \chi_{[-a,a]}(x) e^{-i\xi x} dx = \int_{-a}^{a} e^{-i\xi x} dx = \left[ \frac{1}{-i\xi} e^{-i\xi x} \right]_{-a}^{a} = \frac{e^{-i\xi a}-e^{i\xi a}}{-i\xi} = 2 \frac{\sin a\xi}{\xi} $$

Answer 2

Use three things :

Taylor development of $\phi$, convergence of Dirichlet integral, and Riemann Lesbegues lemma.

Indeed :

$$ \phi(x)=\phi(0)+ x\psi(x) $$

With $\psi$ is in $\mathcal{C}_{Supp Compact} ^\infty $

So your integral :

$$ I_t=\int_R \dfrac{\sin(xt) \phi(0)} {xt} d(xt) + \int_R \dfrac{\sin(xt)x\psi(x) }{x} dx$$

The second integral is null by Riemann-Lesbegues at the infinity. The first is 2 times Dirichlet integral i. e. $\pi$.

$$ I_{\infty} = \pi \phi(0) $$

Thus, your distribution $T_t$ converges so that :

$$ T_t \to \pi \delta_0 $$

Answer 3

Here is a solution that uses known facts about Calculus (Taylor series, integration by parts and Riemann integration, improper integrals) and nothing else.

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We show that for any $\phi\in\mathcal{D}(\mathbb{R})$, $u_t(\phi)=\int_{\mathbb{R}}\frac{\sin(tx)}{x}\phi(x)\,dx\xrightarrow{t\rightarrow\infty}\phi(0)\pi$

Suppose $\operatorname{supp}(\phi)\subset [-A,A]$. Then \begin{aligned} u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi(x)=\int^A_{-A}\frac{\sin tx}{x}\phi(-x)dx \end{aligned} and so, $$ u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi_e(x)\,dx $$ where $\phi_e(x)=\frac12(\phi(x)+\phi(-x))$ is the even part of $\phi$. The advantage of working with $\phi_e$ is that not only is $\phi_e\in\mathcal{D}(\mathbb{R})$, but also $\phi_e(0)=\phi(0)$ and $\phi'_e(0)=0$. By Taylor's theorem

1. $\phi_e(x)=\phi(0)+O(x^2)$ around $x=0$.
2. $\phi'_e(x)=O(x)$ around $x=0$.

With this in mind, we have that $$ u_t(\phi)=\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx +\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx $$ By (1) and (2), the map $\psi(x)=\frac{\phi_e(x)-\phi(0)}{x}$, $x\neq0$ and $\psi(0)=0$, is continuously differentiable. Integrating by parts we obtain $$ \int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx=\frac1t\int^A_{-A}\cos(xt)\Big(\frac{\phi'_e(x)}{x} -\frac{\phi_e(x)-\phi(0)}{x^2}\Big)\,dx $$ As $\phi'_e(x)/x$ and $(\phi_e(x)-\phi(0))/x^2$ are integrable ( Riemann integrable and thus, Lebesgue integrable) over $[-A,A]$, $$ \Big|\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx\Big|\leq\frac{1}{t}\left(\int^A_{-A}\Big|\frac{\phi'_e(x)}{x}\Big|+\Big|\frac{(\phi_e(x)-\phi(0)}{x^2}\Big|\,dx\right)\xrightarrow{t\rightarrow\infty}0 $$ Putting this together, we obtain that $\lim_{t\rightarrow\infty}u_t(\phi)$ exists and $$\lim_{t\rightarrow\infty}u_t(\phi)=\lim_{t\rightarrow\infty}\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx=\lim_{t\rightarrow\infty}\phi(0)\int^{tA}_{-tA}\frac{\sin x}{x}\,dx=\phi(0)\pi$$ That is, $u_t\xrightarrow{t\rightarrow\infty}\pi\delta_0$ in distribution.