Show that $\int_{-\infty}^{+\infty} e^{-(\frac{a}{x^2}+x^2 )}dx=\sqrt{\pi}e^{-2\sqrt{a} }, \forall a>0$.
I know this can be formulated in terms of conditional characteristic function of Cauchy distribution. But that's why I'm here to ask how this integral is computed in the first place (possibly using some complex analysis tools? I'm interested to know!). I tried the standard trick: $$\begin{aligned}\left(\int_{-\infty}^{+\infty} e^{-(\frac{a}{x^2}+x^2 )}dx\right)^2=&\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-\left(\frac{a}{x^2}+\frac{a}{y^2}+x^2+y^2 \right)}dxdy\\=&\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-\left(\frac{a(x^2+y^2)}{x^2y^2}+x^2+y^2 \right)}dxdy\\=&\int_{0}^{2\pi}\int_{0}^{+\infty} e^{-\left(\frac{4a}{\sin^2(2\theta)}+r^2 \right)} rdr d\theta \\=&\int_{0}^{+\infty} e^{-r^2} rdr \int_{0}^{2\pi} e^{-\frac{4a}{ (\sin^2(2\theta))} }d\theta\\=&\int_{0}^{\pi} e^{-\frac{4a}{ (\sin^2(\theta))} }d\theta \end{aligned}$$
How should I continue from here?
