The inverse Fourier transform is defined as:
$$\mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g(k) e^{i k x} d k$$
I can't get an inverse Fourier Transform to
Q1: $$g(k)=\cos (\sqrt{ k^2 d -a^2 })$$ Q2:
$$g(k)=\frac{1}{ \sqrt{ k^2 d -a^2 }}\sin (\sqrt{ k^2 d -a^2 })$$
I would really appreciate some any help .
Q1: $$ \mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \cos (\sqrt{ k^2 d -a^2 }) e^{i k x} d k $$ I see no reason to think it may be written any simpler than that.
Also note that the integrand involves cosine of an imaginary argument, at least on part of the range.