Solve Langevin equation with blue noise?

Question

I would like to solve the following Langevin equation $$\frac{d^2 x}{d t^2}+\omega_0^2x(t)=\eta(t),$$ where $\eta(t)$ is a blue noise signal given by $$\eta(t)=\int_{-\infty}^\infty \hat{\eta}(f)\exp(2\pi i t f) df,$$ and $\hat{\eta}(t)$ is the fourier transform given by $$\hat{\eta}(f)=\int_{-\infty}^\infty \eta(t)\exp(-2\pi i t f) dt,$$ such that $|\hat{\eta}(f)|=D\sqrt{|f|}$, where $D$ is a constant which gives a measure of the strength of the signal. According to Wikipedia the autocorrelation is given by $$\begin{aligned} R_{\eta \eta}(\tau)&=\int_{-\infty}^\infty|\hat{\eta}(f)|^2\exp(2\pi i \tau f) df,\\ &= \frac{D^2}{2\pi^2\tau^2}, \end{aligned}$$ where the Fourier transform is calculated using an identity in Wikipedia. This post shows that the variance is given by $$\begin{aligned} \langle x^2(t) \rangle &= \frac{1}{\omega_0^2}\int_0^t\int_0^t \sin[\omega_0(t-t')]\sin[\omega_0(t-t'')]R_{\eta\eta}(t'-t'')dt''dt' \\ &= \frac{D^2}{2\pi\omega_0^2}\int_0^t\int_0^t \frac{\sin[\omega_0(t-t')]\sin[\omega_0(t-t'')]}{(t'-t'')^2}dt''dt'. \end{aligned}$$ This integral is not defined at $t'=t''$, does this mean we cannot calculate the solution for the case where $\eta(t)$ is blue noise or have I made a mistake in the maths?