How can I compute this limit I want to prove that
$\lim_{h\rightarrow\infty}\left(\int_{0}^{\infty}\left(\cos ht-1\right)\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt\right)=-\int_{0}^{\infty}\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]dt$
where 
$\underset{t}{\triangle}\eta(t)=\eta(t)+\eta(-t)$ and $\phi$ is an integrable function (in the lebesgue sense), to be precise it is the fourier transform of an integrable density function and thus continuous. Also $\phi$ is differentiable at $0$.  
According to the authors of this paper (see proof of theorem 3), this can be achieved by showing $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is integrable and the result will follow from the Riemann Lebesgue lemma. 
They do this by showing that $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded. And this is the part of the proof I am stuck on. Can anyone show me how to prove $\underset{t}{\triangle}\left[\frac{\phi(t)\exp\left(-itx\right)}{it}\right]$ is uniformly bounded and integrable? 
Thanks
 A: The proof of Theorem 3 is discussing $\int_0^\infty exp(-itx) \phi(t) \frac{\cos(ht)-1}{t}\, dt$.  That -1 makes a big difference.
A: I hope I'm not misunderstanding the question, but here's what I think (I apologize ahead of time as it is late!)
Write $\eta(t) = \displaystyle\mathop{\Delta}_{t} \left[\frac{\phi(t) e^{-itx}}{it} \right] = \frac{\phi(t) e^{-itx} - \phi(-t) e^{itx}}{it}$.
Then,
$$
\begin{array}
\\
\int_0^{\infty} (cos(ht) - 1) \eta(t) dt & = \left.\left( \frac{1}{h} \sin(ht) - t\right) \eta(t)\right|_0^{\infty} + \int_0^{\infty} \left(1 - \frac{\sin(ht)}{ht}\right)t \eta'(t) dt \\
&= \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta'(t) dt
\end{array}
$$
 by parts (which is justified as the author assumes $\phi$ is differentiable at the origin).  The Dominated Convergence Theorem implies that
$$
\lim_{h \to \infty} \int_0^{\infty} \left( 1 - \frac{\sin(ht)}{ht} \right) t \eta'(t) dt = \int_0^{\infty} t \eta'(t) dt.
$$
Again, by parts, we have
$$
\int_0^{\infty} t \eta'(t) dt = \left. t \eta(t)\right|_0^{\infty} - \int_0^{\infty} \eta(t) dt = - \int_0^{\infty} \eta(t) dt.
$$
Putting it altogether gives:
$$
\int_0^{\infty} \left( \cos(ht) - 1\right) \eta(t) dt = - \int_0^{\infty} \eta(t) dt.
$$
