Second derivative of $\int_\mathbb{R}\cos(tx)dp(x)$ Let $p$ be a probability on $\mathbb{R}$ and 
$$f(t):=\int_\mathbb{R}\cos(tx)dp(x).$$
I want to show that if $f''(0)$ exists then $$f''(0)=\lim_{t\to 0}2\frac{f(t)-1}{t^2} \: \:(\star).$$
By differentiation under the integral sign, I obtain
$$f''(t)=\int_\mathbb{R}-x^2\cos(tx)dp(x).$$
Now, it seems that I need a sort of "integration by parts for probabilities", but I don't know if something similar exists.
So what's the fastest way to deduce $(\star)$?
 A: Observe that  $f(0)=1$ and $f'(0)=0$. Using Taylor in $0$ we have 
$$f(t)=f(0)+f'(0)t+\frac{f''(0)}{2}t^2+o(t^2).$$
Therefore $$f''(0)=2\frac{f(t)-f(0)-f'(0)t}{t^2}+o(1)=2\frac{f(t)-1}{t^2}+o(1).$$
A: 1) Using Lebesgue's dominated convergence theorem (LDCT), you may commute $\lim \limits _{t \to 0}$ and $\int \limits _{\Bbb R}$, so that
$$\lim \limits _{t \to 0} \big( f(t) - 1 \big) = \lim \limits _{t \to 0} f(t) - 1 = \lim \limits _{t \to 0} \int \limits _{\Bbb R} \cos (tx) \Bbb d p - 1 = \int \limits _{\Bbb R} \lim \limits _{t \to 0} \cos (tx) \Bbb d p - 1 = \int \limits _{\Bbb R} 1 \Bbb d p - 1 = 0 ,$$
therefore one may apply L'Hospital's theorem to obtain $\lim \limits _{t \to 0} 2 \dfrac {f(t) - 1} {t^2} = \lim \limits _{t \to 0} \dfrac {f'(t)} t$.
2) We must now make the assumption that the identity function $x : \Bbb R \to \Bbb R$ is Lebesgue-integrable, or else I suspect your claim to be false!
Using LDCT a second time, one may show that the derivative commutes with the integral, i.e $f'(t) = \dfrac {\Bbb d} {\Bbb d t} \int \limits _{\Bbb R} \cos (tx) \Bbb d p = \int \limits _{\Bbb R} \dfrac {\Bbb d} {\Bbb d t} \cos (tx) \Bbb d p = \int \limits _{\Bbb R} -x \sin (tx) \Bbb d p$, essentially because $| \int \limits _{\Bbb R} -x \sin (tx) \Bbb d p | \le \int \limits _{\Bbb R} |-x \sin (tx)| \Bbb d p \le \int \limits _{\Bbb R} |x|\Bbb d p < \infty$ by the afore-mentioned assumption.
3) Using LDCT a third time like in step 1), it is easy to show that $\lim \limits _{t \to t_0} f'(t) = f'(t_0) \ \forall t_0 \in \Bbb R$ (i.e. that $f'$ is continuous). In particular, $\lim \limits _{t \to 0} f'(t) = f'(0) = 0$. Therefore, one may apply L'Hospital's theorem a second time to get $\lim \limits _{t \to 0} \dfrac {f'(t)} t = \lim \limits _{t \to 0} \dfrac {f''(t)} 1 = f''(0)$.
We have obtained that $f''(0)$ exists and is equal to $\lim \limits _{t \to 0} 2 \dfrac {f(t) - 1} {t^2}$.
