The probability density function of a symmetric Lévy alpha-stable distribution is the inverse Fourier transform of characteristic function $\phi(t)=\exp\left(it\mu-\left|ct\right|^{\alpha}\right)$:
$$f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} \phi(t)e^{-ixt}dt.$$
This yields,
$$\int_{-\infty}^{\infty} \exp\left(i t(\mu-x)-\left|ct\right|^{\alpha}\right)dt=\int_{-\infty}^{\infty} e^{-\left|t\right|^{\alpha}}\left(\cos(t(\mu-x))+i\sin(t( mu-x)\right)dt.$$
Taking the real part of the integrand seems to produce the correct distribution. So, when $\alpha,c=1$,
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\left|t\right|}\cos(t(\mu-x))dx=\frac{1}{\pi\left(\mu^{2}-2\mu x+x^{2}+1\right)}.$$
Integrating this over the support produces $1$ and it is symmetric about $u$, as expected. Is there any reason why ignoring the complex term of the second equation produces the PDF? Also, are there any other values for $\alpha$ which are analytically expressible? Any help would be much appreciated.