For the probability density function in the form:$$f_X(x)=0.5\delta(x+\mu)+0.5N(\mu,\sigma^2)$$ Find the differential entropy defined by $h(X)=-\int_{-\infty}^\infty f_X(x)\log f_X(x)dx$.
Question
Is the follwing solution is valid?
$$h(X)=-\int_{-\infty}^{\infty}f_X(x)\log(f_X(x))dx \\=-\int_{-\infty}^{\infty}0.5\bigg[\delta(x+\mu)+N(\mu,\sigma^2)\bigg]\log(0.5\bigg[\delta(x+\mu)+N(\mu,\sigma^2)\bigg]) \\ \\=-0.5\int_{-\infty}^{\infty}\bigg[\delta(x+\mu)+N(\mu,\sigma^2)\bigg]\log(0.5) \\-0.5\int_{-\infty}^{\infty}\delta(x+\mu)\log\bigg[\delta(x+\mu)+N(\mu,\sigma^2)\bigg] \\-0.5\int_{-\infty}^{\infty}N(\mu,\sigma^2)\log\bigg[\delta(x+\mu)+N(\mu,\sigma^2)\bigg] \\=-\log(0.5)-0.5\log\bigg[1+N(\mu,\sigma^2)\bigg]_{x=-\mu} \\-0.5N(\mu,\sigma^2)\log\bigg[1+N(\mu,\sigma^2)\bigg]_{x=-\mu} \\-0.5\int_{-\infty}^{\infty}N(\mu,\sigma^2)\log\bigg[N(\mu,\sigma^2)\bigg] \\+0.5N(\mu,\sigma^2)\log\bigg[N(\mu,\sigma^2)\bigg]_{x=-\mu}$$
I think you got muddled up and mixed the random variables with their probability density functions. I guess that you should have a random variable $X \sim \frac{1}{2}\mathrm{Delta}(-\mu) + \frac{1}{2}\mathcal{N}(\mu,\sigma^2)$, the entropy of which you would like to compute.
As Dirac delta distributions are deterministic in nature, it behaves as a constant and, in the present case, centers the adjacent normal law, i.e. $X \equiv \frac{1}{2}\left(-\mu+\mathcal{N}(\mu,\sigma^2)\right) = \mathcal{N}(0,\sigma^2/4)$ whose entropy is simply given by $H[X] = H[\mathcal{N}(0,\sigma^2/4)] = \ln\sqrt{2\pi e} + \ln(\sigma^2/4)$.