Distrubtional limit of $f_i(x)=i\sin(i|x|)$ as $i\to\infty$

Question

Please help me solve the following:

For $i \in \mathbb{N}$, let $$f_i(x)=i\sin(i|x|)$$ find the distributional limit as $i \to \infty$.

Answer 1

We split the integral into two by $\int_{-\infty}^{\infty} = \int_{-\infty}^{0} + \int_{0}^{\infty}$. An integration by parts of the integral on the positive $x$-axis gives $$ \int_{0}^{\infty} n \sin(nx) \, \varphi(x) \, dx = [-\cos(nx) \, \varphi(x)]_{0}^{\infty} - \int_{0}^{\infty} (-\cos(nx)) \, \varphi'(x) \, dx \\ = \varphi(0) + \int_{0}^{\infty} \cos(nx) \, \varphi'(x) \, dx . $$ Similarily, an integration by parts of the integral on the negative $x$-axis gives $$ \int_{-\infty}^{0} n \sin(-nx) \, \varphi(x) \, dx = \{ x \to -x \} = \int_{\infty}^{0} n \sin(nx) \, \varphi(-x) \, (-dx) = \int_{0}^{\infty} n \sin(nx) \, \varphi(-x) \, dx \\ = \varphi(0) - \int_{0}^{\infty} \cos(nx) \, \varphi'(-x) \, dx . $$

Together we get $$ \int_{-\infty}^{\infty} n \sin(n|x|) \, \varphi(x) \, dx = \int_{0}^{\infty} n \sin(nx) \, \varphi(x) \, dx + \int_{-\infty}^{0} n \sin(-nx) \, \varphi(x) \, dx \\ = 2 \varphi(0) + \int_{0}^{\infty} \cos(nx) \, (\varphi'(x)-\varphi'(-x)) \, dx . $$

Here, $$ \left|\int_{0}^{\infty} \cos(nx) \, (\varphi'(x)-\varphi'(-x)) \, dx\right| \\ = \left| [\frac{1}{n}\sin(nx) (\varphi'(x)-\varphi'(-x))]_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{n}\sin(nx) \, (\varphi''(x)+\varphi''(-x)) \, dx \right| \\ = \left| \frac{1}{n} \int_{0}^{\infty} \sin(nx) \, (\varphi''(x)+\varphi''(-x)) \, dx \right| \leq \frac{1}{n} \int_{0}^{\infty} |\varphi''(x)+\varphi''(-x)| \, dx \\ \rightarrow 0 . $$

Thus, $n \sin(n|x|) \rightarrow 2\delta(x)$ as a distribution.