Let $\mu$ be a Gaussian measure on a Hilbert space $(U, \mathscr{B}(U))$. Suppose that we have a symmetric positive definite bounded linear operator $Q \in L(U)$ defined by $\langle Q h_1, h_2 \rangle_U = \int \langle x, h_1 \rangle_U \langle x , h_2 \rangle_U \mu(dx)$ (we assume $\mu$ has mean zero). I would like to prove that $Q$ is trace class. In the proof there is an equality given by $$1-e^{-\frac{1}{2} \langle Qh, h\rangle_U} = \int_U (1-\cos \langle h , x\rangle_U )\mu(dx).$$However, I cannot understand how to get this equality. I would greatly appreciate if anyone could explain this to me.
This is the part of the notes I'm referring to.
Let $m$ be the mean of $\mu$. The characteristic function of Gaussian measure is given by $$\phi (h) = \mathbb{E}(\exp (i \langle h, x\rangle_U)) = \exp \left( i \langle h, m \rangle_U - \frac{1}{2}\langle Qh, h\rangle_U\right) \underbrace{=}_{m = 0} \exp\left(- \frac{1}{2}\langle Qh, h\rangle_U\right)$$
By Euler's identity $$\phi (h) = \mathbb{E}(\exp (i \langle h, x\rangle_U)) = \mathbb{E}(\cos ( \langle h, x\rangle_U))+ i \mathbb{E}(\sin ( \langle h, x\rangle_U))$$
But since $\phi (h)$ is purely real when $m = 0$, we have $\mathbb{E}(\sin ( \langle h, x\rangle_U)) = 0$, which implies that $$\exp\left(- \frac{1}{2}\langle Qh, h\rangle_U\right) = \mathbb{E}(\cos ( \langle h, x\rangle_U))$$ Translating this result to your problem, we get $$\int_U 1-\cos ( \langle h, x\rangle_U) \mu(dx) = \mathbb{E}(1-\cos ( \langle h, x\rangle_U)) = 1 - \exp\left(- \frac{1}{2}\langle Qh, h\rangle_U\right)$$ As desired.