$$\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos qx\,dq$$ $$=\frac{1}{\pi}\gamma^{-1/\alpha} \Gamma \left(1+\frac{1}{\alpha}\right)-\frac{1}{2\pi\alpha}x^2 \gamma^{-3/\alpha} \Gamma \left(\frac{3}{\alpha}\right)+{\cal O}(x^4).$$

$$P_{\alpha,\gamma}(x)=\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos qx\,dq$$ $$=\frac{1}{\pi}\gamma^{-1/\alpha} \Gamma \left(1+\frac{1}{\alpha}\right)-\frac{1}{2\pi\alpha}x^2 \gamma^{-3/\alpha} \Gamma \left(\frac{3}{\alpha}\right)+{\cal O}(x^4).$$

To check, I take $\alpha=1$ or $\alpha=2$, when $$P_{1,\gamma}(x)=\frac{\gamma}{\gamma^2+x^2}=\frac{1}{\gamma}-\frac{x^2}{\gamma^3}+{\cal O}(x^4),$$ $$P_{2,\gamma}(x)=\frac{\sqrt{\pi } e^{-\frac{x^2}{4 \gamma}}}{2 \sqrt{\gamma}}=\frac{\sqrt{\pi }}{2 \sqrt{\gamma}}-\frac{\sqrt{\pi } x^2}{8 \gamma^{3/2}}+{\cal O}(x^4),$$ in agreement with the expansion above.

$$\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos(qx)\,dq$$ $$=\frac{1}{\pi}\gamma^{-1/\alpha} \Gamma \left(1+\frac{1}{\alpha}\right)-\frac{x^2 \gamma^{-3/\alpha} \Gamma \left(\frac{3}{\alpha}\right)}{2\pi \alpha}+{\cal O}(x^4).$$

$$\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos qx\,dq$$ $$=\frac{1}{\pi}\gamma^{-1/\alpha} \Gamma \left(1+\frac{1}{\alpha}\right)-\frac{1}{2\pi\alpha}x^2 \gamma^{-3/\alpha} \Gamma \left(\frac{3}{\alpha}\right)+{\cal O}(x^4).$$