PSD and $\lim_{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T x(t)\bar y(t)\,dt$
For power signals $x(t)$ and $y(t)$, the function

$$R_{xy}(\tau)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(t)\bar{y}(t+\tau)dt\tag{1}$$

is the cross-correlation of $x(t)$ and $y(t)$. So the expression you're asking about is the cross-correlation of $x(t)$ and $y(t)$ evaluated at lag $\tau=0$:

$$R_{xy}(0)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(t)\bar{y}(t)dt\tag{2}$$

Since the Fourier transform of $(1)$ is the cross-power spectrum $S_{xy}(\omega)$ of $x(t)$ and $y(t)$, the expression in $(2)$ is related to $S_{xy}(\omega)$ by:

$$R_{xy}(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{xy}(\omega)d\omega\tag{3}$$

So the expression in your question is proportional to the integral of the cross-power spectrum of the power signals $x(t)$ and $y(t)$.