My question relates to the evaluation of the singular integral for Waring's problem, following pp. 460 of Iwaniec and Kowalski's "Analytic Number Theory".
Background: There, we are interested in evaluating the integral $$V_f(B) := \int_{-\infty}^\infty \int_B e(\theta f(x))\,\mathrm{d}x\,\mathrm{d}\theta,$$ where $B\subset\mathbb{R}^s$ is some box and $f(x)$ is some polynomial (and we write $e(t):=e^{2\pi it}$). I am particularly interested in the case when $f(x)=N-x_1^k-\cdots - x_s^k$ and $B=[0,N^{1/k}]^s$.
Iwaniec and Kowalski introduce a factor of $\frac{\sin(2\pi\theta V)}{\pi\theta}$ in order to make use of the fact that $$\int_{-\infty}^\infty \frac{\sin(2\pi\theta V)}{\pi\theta} e(\theta y)\,\mathrm{d}\theta = \begin{cases} 1 & |y|<V \\ 0 & |y|>V \end{cases}$$ and conclude that for any $V>0$, (with some acceptable error term, which I omit here) $$V_f(B) = \int_{-\infty}^\infty \frac{\sin(2\pi\theta V)}{2\pi\theta V} \int_B e(\theta f(x))\,\mathrm{d}x\,\mathrm{d}\theta = \frac{1}{2V} \lambda(\{x\in B : |f(x)|\leq V\}),$$ where $\lambda$ is Lebesgue measure.
My problem: I think I can obtain this if I can exchange the order of integration, since then $$V_f(B) = \frac{1}{2V}\int_B \int_{-\infty}^\infty \frac{\sin(2\pi\theta V)}{\pi\theta} e(\theta f(x))\,\mathrm{d} \theta\,\mathrm{d}x = \frac{1}{2V} \int_B 1_{|f(x)|\leq V}\,\mathrm{d}x.$$ However, I am struggling to justify the exchange of order of integration because Fubini-Tonelli fails here, since $\frac{|\sin(2\pi\theta V)|}{|\pi\theta|}$ is not integrable on $(-\infty,\infty)$. I have thought about the fact that $$V_f(B) = \frac{1}{2V} \lim_{T\to\infty} \int_{-T}^T \int_B \frac{\sin(2\pi\theta V)}{\pi\theta} e(\theta f(x))\,\mathrm{d}x\,\mathrm{d}\theta = \frac{1}{2V} \lim_{T\to\infty} \int_B \int_{-T}^T \frac{\sin(2\pi\theta V)}{\pi\theta} e(\theta f(x))\,\mathrm{d}\theta\,\mathrm{d}x,$$ from which I think it would be enough to show that $$ F_m(x) := \int_{-m}^m \frac{\sin(2\pi\theta V)}{\pi\theta} e(\theta f(x))\,\mathrm{d}\theta$$ is dominated by some integrable function $g(x)$, so that dominated convergence allows taking the limit inside the integral over $B$. However I cannot find such an integrable function uniformly in $m$.
How might I justify the exchange of order of integration? Or is there some other way to see that the result is true?