Find $\lim_{\epsilon\to 0}\frac{1}{\epsilon^3}\int\limits_{0}^{\infty}\sin(\epsilon\cdot f(t))d\mu(t)$ given $\int\limits_{0}^{\infty}f=0$ Let $(\mathbb {R},\mathcal B, \mu)$ be finite measure space, where $\mathcal B$ is the Borel sigma-algebra on $\mathbb R$.
Let $f$ be an integrable function such that $$ \int\limits_{0}^{\infty}f(x)d\mu = 0.$$
Also assume $f^2$ is integrable, i.e $\int\limits_\mathbb {R} f^2d\mu<\infty.$
find $$\lim_{\epsilon \to 0}\frac{1}{\epsilon ^ 3}\int\limits_{0}^{\infty}\sin(\epsilon\cdot f(t))d\mu(t)$$ 
 A: If $f\in L^4$ then the answer is $-\frac{1}{6} \int_0^{\infty} f^3 d\mu$. There is consequently no consistent answer when we only know that $f^2$ is integrable.
Indeed, if $f\in L^4$ then write
$$ \sin\left(\epsilon f\left(t\right) \right) = \epsilon f\left(t\right) - \frac{1}{6} \epsilon^3 f\left(t\right)^3 + \frac{\sin\left(c_t\right)}{24} \epsilon^4 f\left(t\right)^4, $$
and integrate:
$$ \frac{1}{\epsilon^3} \int_0^{\infty} \sin\left(\epsilon f\left(t\right)\right) d\mu = -\frac{1}{6} \int_0^{\infty} f^3 d\mu + \frac{\epsilon}{24} \int_0^{\infty} \sin\left(c_t\right)f^4\left(t\right) d\mu\left(t\right)  . $$
The error term (i.e. the last term on the right) can be bounded as follows, using $\left|\sin\left(c_t\right)\right| \leq 1$:
$$ \frac{\epsilon}{24} \left|\int_0^{\infty} \sin\left(c_t\right)f^4\left(t\right) d\mu\left(t\right)\right| \leq \left(\frac{\int_0^{\infty} f^4 d\mu}{24}\right) \epsilon \xrightarrow[\epsilon\searrow 0]{}0, $$
since $f^4$ is integrable.
A: The limit may be $+\infty$. Let $\mu$ be the Lebesgue measure. Let $f(t) = \epsilon_t\frac{1}{t^{.49}}$ where $\epsilon_t = 0$ for $t \le 0$ and $t \ge 1$, $\epsilon_t = 1$ for $0 < t \le t_*$, and $\epsilon_t = -1$ for $t_* < t < 1$, where $t_* = \frac{1}{2^{100/51}} \approx .25689$ is such that $\int_\mathbb{R} f(t)dt = 0$. Take $\epsilon > 0$ very small. Then, for $t < \epsilon^{1/.49}$, we have $\sin(\epsilon\frac{\epsilon_t}{t^{.49}}) \sim \epsilon\frac{\epsilon_t}{t^{.49}}$, so $\int_{\epsilon^{1/.49}}^1 \sin(\epsilon f(t))dt \sim \int_{\epsilon^{1/.49}}^1 \epsilon\frac{\epsilon_t}{t^{.49}}dt = -\epsilon\int_0^{\epsilon^{1/.49}} \frac{1}{t^{.49}}dt = -\frac{100}{51}\epsilon^{100/49}$. 
Ok, I actually have to go now, but the leftover region is of size $\epsilon^{1/.49} = \omega(\epsilon^3)$, which is bad. The situation for $f(t) = \epsilon_t\frac{1}{t^{1/8}}$ is very different, since then $\sin(\epsilon f(t)) \sim \epsilon f(t)$ for all $t > \epsilon^8$, so the leftover error is at most $\epsilon^8$, which is small enough. 
