$\int^{\infty}_{-\infty}u(x,y) \,d y$ independent of x

Question

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 $\frac{1}{2\pi}$.

Answer 1

Are you sure that there is no typo in your equation ?

With : \begin{eqnarray} 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) \end{eqnarray}

I do not find $I=\int^{\infty}_{-\infty}u(x,y) \,d y \simeq 2.38$ and $I(x)$ is not constant.

For example : Case $x=0$ :

$$u(0,y) = \frac{1}{2\pi}\exp\left(-y^2/2\right)K_0\left(\sqrt{(-y)^2+(y^2/2)^2}\right)$$ $$I = \int^{\infty}_{-\infty}u(0,y) \,d y \simeq 0.64965$$ Note : this numerical result is not valid because some precautions where not taken at that time. More recent numerical results confirmes the value $0.378936$ given by chatur.

In the radical, are the powers not on the same order ?

Note :

The numerical computation is hazarduous because $K_0(0)=\infty$.

The integral is not convergent in the usual sens. But it can be convergent if we consider the Cauchy principal value.

During numerical computation, $y$ varies from a low value $<x$ to an high value $>x$. Proceeding by steps, it can happend that $y$ comes very close to $x$ and the argument of $K_0$ be very close to $0$. In that case, the numerical calculus involves transitorily some differences between very big numbers.

All depends how the sofware detect the singular point and how he manage to treat it as a Cauchy principal value integration. I do not know how Mathematica proceed. Presently, I cannot say if the numerical values obtained are significant or not. If it was poved that the results from Mathematica are correct, I would take my hat off !

NOTE 2 :

As pointed out by GEdgard, the singularity at $x=y$ i.e. at $K_0(0)$ is logarithmic. So, there is no major difficulty for numerical computation, in so far some precautions are taken. I made a few numerical tests on this point.

What is more, for several values of $x$ , I computed the derivative $\frac{dI}{dx}$, which is the integral of $\frac{du}{dx}$ . The results are very close to $0$.

This draw to think that the chatur's conjecture $I(x)=$constant might be exact. Of course, this is note a prove.

Answer 2

I think there is a typo somewhere. I tell you why. First of allI write your equation as folllow \begin{equation} u(x,y)=g(x,y) K_0(\rho(x,y)) \end{equation} where \begin{equation} g(x,y) =\frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right) \end{equation} And \begin{equation} \rho(x,y) =\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2} \end{equation} Now please note that when differentiating $g(x,y)$ with respect to $x$ we have \begin{equation} g_x(x,y) =x g(x,y) \end{equation} Then \begin{equation} \frac{d}{dx} u(x,y)=g(x,y) \left(x K_0(\rho(x,y))-K_1(\rho(x,y)) \rho_x(x,y)\right)=0 \end{equation} Differentiating $u(x,y)$ a second time we have: \begin{equation} \frac{d}{dx} u_x(x,y)=g_x(x,y) \left(x K_0(\rho(x,y))-K_1(\rho(x,y)) \right)+g(x,y) \left(K_0(\rho(x,y))-x K_1(\rho(x,y)) \rho_x(x,y)- K_1(\rho(x,y)) \rho_{xx}(x,y) +\frac{\rho^2_x(x,y)}{2} \left(K_0(\rho(x,y))+K_2(\rho(x,y)) \right)\right )=0 \end{equation} Setting the first and second derivative to 0, using the equation $g_x(x,y)= x g(x,y)$ and the property of the modified Bessel function of second kind: \begin{equation} \rho(x,y) \left(K_2(\rho(x,y))-K_0(\rho(x,y)) \right)=2 K_1(\rho(x,y)) \end{equation} It is possible to write the following expression: \begin{equation} \rho(x,y) \rho_x^3(x,y)+x \rho_x^2(x,y)+ \rho_x(x,y) \rho(x,y)(1-x^2)-\rho_{xx}\rho(x,y)=0 \end{equation} This equation must be satisfied by $\rho(x,y)$ defined above. If I dind't do any error (please check because I filled few pages and I'm not completely sure) the $\rho(x,y)$ as define above doesn't satisfy the differential equation for $\rho$.

Answer 3

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.

$$\implies I = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{1}{4s}\right)}{\sqrt{s} (1+4s)} \, ds\stackrel{s\,\mapsto\frac{1}{s}}{=}\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\exp\left(-\frac{s}{4}\right)}{\sqrt{s} (4+s)} \, ds=\frac{e}{2} \sqrt{\pi} \, \text{erfc} (1)=0.378\cdots$$

$\square$