Let $X(t)$ be a stationary Gaussian process, $EX(t)=0$, the correlation function $R(\tau)$ is given. What bounds from above can be given for the $p$-th moment
($p>0, p \in \mathbb{R}$) of the integral 
$$
 \int\_0^T |X(t)|^2 dt?
$$
(The integral is the pathwise integral.)



Inequalities for moments of a certain integral
*I will be working with the cumulants rather than the moments.*


To simplify the notation let me choose units so that $T\equiv 1$ and discretize time $t\_n=n/N$, collecting $X(t\_1),X(t\_2),\ldots X(t\_N)$ in a vector $\vec{X}$. Hence
$$\int\_0^1 |X(t)|^2 dt\rightarrow N^{-1}\vec{x}\cdot\vec{x}\equiv Z.$$
The correlator $E(\vec{x}\vec{x})=\mathbf{R}$ is an $N\times N$ positive definite Toeplitz matrix: the matrix elements $\mathbf{R}\_{nm}=R(|n-m|)\geq 0$ depend only on the time difference because of the stationarity of the Gaussian process.  

The cumulant generating function is
$$F(\xi)=\log E(e^{\xi Z})=-\tfrac{1}{2}\log\,{\rm det}\,(1-2\xi \mathbf{R})=-\tfrac{1}{2}{\rm tr}\,\log\,(1-2\xi \mathbf{R}).$$
By expanding the logarithm in powers of $\xi$ we obtain the $p$-th cumulant $\mu\_p$ of $Z$ as
$$\mu\_p=2^{p-1}(p-1)!\,{\rm tr}\,\mathbf{R}^p.$$
Reverting back to continuous time, and reinstating the time $T$,
$$\mu\_p=2^{p-1}(p-1)!\int\_0^T dt\_1\int\_0^T dt\_2\cdots \int\_0^T dt\_p\, R(t\_1-t\_2)R(t\_2-t\_3)\cdots R(t\_{p-1}-t\_{p})R(t\_p-t\_1).$$


If the correlator $R(t)$ reaches its maximum at $t=0$, then we have the upper bound
$$\mu\_p\leq 2^{p-1}(p-1)!T^p R(0)^p.$$
An improved upper bound can be obtained if $R(t)/R(0)$ is maximised by a box function $B(x,a)=1$ if $|x|<a$ and $B(x,a)=0$ if $|x|>a$. Then
$$\mu\_p\leq 2^{p-1}(p-1)!T^p R(0)^p f\_p(a).$$
The function $f\_p(a)$ is the convolution of the box function on the $p$-dimensional hypercube. I don't have a closed form expression for general $p$ --- I have asked at [MSE](https://math.stackexchange.com/q/3233160/87355) --- small-$p$ expressions are


$$f\_2(a)=2a-a^2,\;\;f\_3(a)=3a^2-2a^3,$$
$$f\_4(a)=\begin{cases}
\frac{2}{3} \left(-7 a^4+8 a^3\right)&\text{if}\;\;0<a<1/2,\\
\frac{1}{3} \left(2 a^4-16 a^3+24 a^2-8 a+1\right)&\text{if}\;\;1/2<a<1.
\end{cases}$$


