Considering random variable $w$ with the following distribution
$$w\sim\pi\mathcal{N}(0,\alpha^{-1})+(1-\pi)\delta_0$$
where $\pi$ is a random variable with the following distribution
$$\pi\sim\mathcal{B}eta(a_0, b_0)$$
and $\delta_0$ is a point mass at zero, if we would like to integrate the distribution over $w$ such that
$$I=\int_{-\infty}^{\infty} \Big[\pi \big(\frac{1}{2\pi\alpha^{-1}}\big)^{k/2}\exp\big(-\frac{\|w\|^2}{2\alpha^{-1}}\big) +(1-\pi)\delta_0 \Big]dw$$
The second part of the integral would have the form
$$II=\int_{-\infty}^{\infty} \Big[(1-\pi)\delta_0\Big] dw = (1-\pi) \int_{-\infty}^{\infty} \Big[\delta_0\Big] dw = (1-\pi) $$
Now my problem starts if we do the integration w.r.t. $\pi$ rather than $w$, so that the second part become
$$II=\int_{-\infty}^{\infty} \Big[(1-\pi)\delta_0\Big] d\pi,$$
although $\delta_0$ is a function of $w$, the probability distribution of $w$ is a function of $\pi$, so it can’t be taken out of integral and the delta is a function of $w$ rather than $\pi$, so I can't solve it like the previous part.