Evaluate $\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}\,\mathrm dx$ How to integrate 
$$\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}\,\mathrm dx$$
where $t$ is a constant/parameter?
 A: Maybe:
$$\int_0^\infty e^{-x^2}\cos\frac{t}{x^2}dx=\int_0^\infty \frac{1}{2}e^{-x^2}\left(e^\frac{it}{x^2}+e^\frac{-it}{x^2}\right) dx\\
=\int_0^\infty \frac{1}{2}\left(e^{-x^2+\frac{it}{x^2}}+e^{-x^2-\frac{it}{x^2}}\right) dx$$
Complete the square for the second term $\left(x+\sqrt t\frac{1-i}{\sqrt2}x^{-1}\right)^2-\sqrt {2t}i=x^2+itx^{-2}$
And similar for the first term... 
Is it possible to integrate this out with the substitutions $$u=x+(1-i)\sqrt\frac{t}{2}x^{-1}$$ and 
$$s=x+(1+i)\sqrt\frac{t}{2}x^{-1}$$? 
A: Let be
$$
\Bbb{F}(t)=\int_0^{\infty}\textrm{e}^{-x^2}\cos\left(\frac{t}{x^2}\right)\textrm{d}x.
$$
Observing that $\cos\theta=\frac{\textrm{e}^{i\theta}+\textrm{e}^{-i\theta}}{2}$, and putting $x^2=u$, we have
$$
\Bbb{F}(t)=\frac{1}{2}\int_0^{\infty}\left[\textrm{e}^{-x^2+i\frac{t}{x^2}}+\textrm{e}^{-x^2-i\frac{t}{x^2}}\right]\textrm{d}x=\frac{1}{2}\int_0^{\infty}\textrm{e}^{-u+\frac{it}{u}}\frac{\textrm{d}u}{2u^{\frac{1}{2}}}+\frac{1}{2}\int_0^{\infty}\textrm{e}^{-u-i\frac{t}{u}}\frac{\textrm{d}u}{2u^{\frac{1}{2}}}
$$
Recalling the integral representation of the Modified Bessel Function of the Second Kind $K_{\nu}(x)$ we have
$$
\Bbb{F}(t)=\frac{1}{4}\frac{K_{-\frac{1}{2}}(z)}{\frac{1}{2}\left(\frac{z}{2}\right)^{\frac{1}{2}}}+\frac{1}{4}\frac{K_{-\frac{1}{2}}(z^*)}{\frac{1}{2}\left(\frac{z^*}{2}\right)^{\frac{1}{2}}}
$$
where $z=\sqrt{-it4}=\sqrt{2|t|}(1-i)$ and $z^*=\sqrt{2|t|}(1+i)$ is the conjugate of $z$.
Using the expression for $K_{-\frac{1}{2}}(x)=K_{\frac{1}{2}}(x)=\sqrt{\frac{\pi}{2x}}\textrm{e}^{-x}$, we find
$$
\Bbb{F}(t)=\frac{\sqrt{\pi}}{4}\left[\textrm{e}^{-z}+\textrm{e}^{-z^*}\right]=\frac{\sqrt{\pi}}{2}\textrm{e}^{-\sqrt{2|t|}}\cos\left(\sqrt{2|t|}\right).
$$
A: I can give you only the results for different $t$s.
Let $t=1$ be, then 
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{1}{x^2}\right)dx = \frac{e^{-\sqrt{2}}\sqrt{\pi}\cos(\sqrt{2})}{2}$$
If $t=2$ is, then
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{2}{x^2}\right)dx = \frac{e^{-2}\sqrt{\pi}\cos(\sqrt{4})}{2}$$
If $t=3$, then
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{3}{x^2}\right)dx = \frac{e^{-\sqrt{6}}\sqrt{\pi}\cos(\sqrt{6})}{2}$$
So:
Is $t$ a constant, then we have:
$$\int _0^{\infty }e^{-x^2}\cos \left(\frac{t}{x^2}\right)dx = \frac{e^{-\sqrt{2t}}\sqrt{\pi}\cos(\sqrt{2t})}{2}$$
