<s>[INST] $\int^{\infty}_{-\infty}u(x,y) \,d y$ independent of x I need to prove that
$$I = \int^{\infty}_{-\infty}u(x,y) \,dy$$
is independent of $x$ and find its value, where
$$u(x,y) = \frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right)K_0\left(\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2}\right)$$
and $K_0$ is the modified Bessel function of the second kind with order zero.
Evaluating the integral numerically with Mathematica for different values of $x$ gives the result of $2.38$, but I want to know if it is possible to show analytically.
Increasing $x$ results in an increase of the exponential term on the left, but it also then strongly increases the argument of modified Bessel function, thus reducing its value.
To show that integral is independant of $x$, it is sufficient to show that $\int^{\infty}_{-\infty}\frac{\, d}{\, dx}u(x,y) = 0$ but any differentiation looks more and more ugly.
EDIT
Mathematica test:
 x = 100
    NIntegrate[
    (1/(2 Pi))* Exp[x*x/2 - y*y/2] BesselK[0, 
       Sqrt[(x - y)*(x - y) + (x*x/2 - y*y/2)*(x*x/2 - 
            y*y/2 )]], {y, -Infinity, x, Infinity}, MaxRecursion -> 22]

This gives an answer of $0.378936$ independent of the choice of $x$. In the earlier calculation I missed the factor $\frac{1}{2\pi}$.
 [/INST] Although I'm about 7 years late, here is an answer anyway for anyone interested:

Claim
$$I = \frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)$$
and is thus independent of $x$.

Proof.
By https://dlmf.nist.gov/10.32#E10
$$K_0 (z) = \frac{1}{2} \int_{0}^{\infty} \exp \left(-t-\frac{z^2}{4t}\right) \, \frac{dt}{t}$$
This allows us to write
$$\begin{align}I &= \int_{0}^{\infty} u(x,y) \, dy + \int_{-\infty}^{0} u(x,y) \, dy \\&=\frac{1}{2\pi}\int_0^\infty e^{-v^2/2}\int_0^\infty\frac{e^{xv}e^{-t-v^2[1+(x-v/2)^2]/(4t)}+e^{-xv}e^{-t-v^2[1+(x+v/2)^2]/(4t)}}t\,dt\,dv\\ &=\frac{1}{2\pi}\int_0^\infty\int_0^\infty \frac{\cosh\left(xv\left(\frac{v^2}{4t}+1\right)\right)}{t}\exp\left(-t-\frac{1+2t+x^2}{4t}v^2-\frac{v^4}{16t}\right)\,dv\,dt\end{align}$$
We now enforce the substitution $t = s v^2 \implies dt = v^2 \,ds$:
$$\begin{align}I&=\frac{1}{2\pi} \int_{0}^{\infty} \int_{0}^{\infty}\frac{\cosh\left(x v \left(\frac{1}{4s}+1\right)\right)}{s} \exp\left(-sv^2-\frac{1+2sv^2+x^2}{4s}-\frac{v^2}{16s}\right) \, dv \, ds\\&=\frac{1}{2\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cosh\left(xv\left(\frac{1}{4s}+1\right)\right)}{s}\exp\left(-\frac{1+x^2}{4s}\right)\exp\left(-v^2 \frac{(1+4s)^2}{16s}\right)\,dv\,ds\end{align}$$
Since
$$\int_{0}^{\infty}\cosh(a v)\exp\left(-v^2 b^2\right) \, dv=\frac{\sqrt{\pi}}{2b}\exp\left(\frac{a^2}{4b^2}\right)$$
$$\implies I = \frac{1}{2\pi} \int_{0}^{\infty} \frac{\exp \left(-\frac{1+x^2}{4s}\right)}{s} \cdot \frac{\sqrt{\pi}}{2\left(\frac{1+4s}{4\sqrt{s}}\right)}\exp \left(\frac{4s x^2 \left(\frac{1}{4s}+1\right)^2}{(1+4s)^2}\right)\, ds$$
This results in the integral being independent as we wanted since the $x^2$ terms cancel.
$$\implies I = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{1}{4s}\right)}{\sqrt{s} (1+4s)} \, ds\stackrel{s\,\mapsto\frac{1}{s}}{=}\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{s}{4}\right)}{\sqrt{s} (4+s)} \, ds=\frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)=0.378\cdots$$
$\square$
</s>