Distrubtional limit of $f_i(x)=i\sin(i|x|)$ as $i\to\infty$ Please help me solve the following:
For $i \in \mathbb{N}$, let $$f_i(x)=i\sin(i|x|)$$ find the distributional limit as $i \to \infty$.
 A: 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.
