Problem: Let $\mu$ be a probability measure on $\left(\mathbb{R},\mathcal{B}\right)$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$. For $t\in\mathbb{R}^+$, I want to show
$$\int_{\mathbb{R}}x^2\,d\mu(x) = 2\lim_{t\to0}\frac{1-\int_{\mathbb{R}}\cos(tx)\,d\mu(x)}{t^2}.$$
I'm not really sure how to go about doing this.
I know by the Dominated Convergence Theorem
$$\lim_{t\to0}\int_{\mathbb{R}}\cos(tx)\,d\mu(x) = 1.$$
Therefore, distributing the limit we get an indeterminate form $(\frac{0}{0})$. So, applying L'hopital's rule and Leibniz integral rule
$$2\lim_{t\to0}\frac{1-\int_{\mathbb{R}}\cos(tx)\,d\mu(x)}{t^2} = 2\lim_{t\to0}\frac{-\frac{d}{dt}\int_{\mathbb{R}}\cos(tx)\,d\mu(x)}{2t} = \lim_{t\to0}\frac{\int_{\mathbb{R}}x\sin(tx)\,d\mu(x)}{t}.$$
EDIT: If I can again apply the DCT, which I don't know if I can, then the following is what I'd do:
Similarly as before, by the Dominated Convergence Theorem
$$\lim_{t\to0}\int_{\mathbb{R}}x\sin(tx)\,d\mu(x) = 0.$$
Therefore, distributing the limit we get an indeterminate form $(\frac{0}{0})$. So, applying L'hopital's rule and Leibniz integral rule
$$\lim_{t\to0}\frac{\int_{\mathbb{R}}x\sin(tx)\,d\mu(x)}{t} = \lim_{t\to0}\frac{d}{dt}\int_{\mathbb{R}}x\sin(tx)\,d\mu(x) = \lim_{t\to0}\int_{\mathbb{R}}x^2\cos(tx)\,d\mu(x).$$
Lastly, by the Dominated Convergence Theorem again
$$\lim_{t\to0}\int_{\mathbb{R}}x^2\cos(tx)\,d\mu(x) = \int_{\mathbb{R}}x^2\,d\mu(x).$$
Though, I'm not sure where to go from here, or even if this approach is correct. Perhaps writing the $2$, in front of the limit, as integral may be helpful, i.e.
$$2 = \int_{\mathbb{R}}2\,d\mu(x).$$
Any help is appreciated.
We have $$ 2\times\frac{1-\int_{\mathbb R}\cos(tx)\,\mu(dx)}{t^2}=2\int_{\mathbb R}\frac{1-\cos(tx)}{t^2}\,\mu(dx). $$
Suppose first that $\int_{\mathbb R}x^2\,\mu(dx)<+\infty$. Since $$ 2\times\frac{1-\cos(tx)}{t^2}\underset{t\to0}{\longrightarrow}x^2, $$ and $$ 0\le2\times\frac{1-\cos(tx)}{t^2}\le x^2, $$ we conclude by the dominated convergence theorem.
Suppose now that $\int_{\mathbb R}x^2\,\mu(dx)=+\infty$. Then by Fatou's lemma, $$ +\infty=\int_{\mathbb R}x^2\,\mu(dx)=2\int_{\mathbb R}\liminf_{t\to0}\frac{1-\cos(tx)}{t^2}\,\mu(dx)\le2\liminf_{t\to0}\int_{\mathbb R}\frac{1-\cos(tx)}{t^2}\,\mu(dx), $$ hence $$ 2\int_{\mathbb R}\frac{1-\cos(tx)}{t^2}\,\mu(dx)\underset{t\to0}{\longrightarrow}+\infty=\int_{\mathbb R}x^2\,\mu(dx). $$