Solution to the “near-Gaussian” integral $\int_{0}^{\infty} e^{- \Lambda \sqrt{(z^2+a)^2+b^2}}\mathrm{d}z$

In the course of my research I have come across the following integral:

$\int_{0}^{\infty} e^{- \Lambda \sqrt{(z^2+a)^2+b^2}}\mathrm{d}z$.

This initially looks like it should be solvable by some suitable change of variable which will allow you to get it into a gaussian form. Unfortunately after trying for awhile I cannot find one. The constants $a$ and $b$ are combinations of parameters $s \in (0,\infty)$, $x \in [0,\infty)$:

$a = s^2-x^2$ and $b = 2sx$.

So the integral can be rewritten as:

$\int_{0}^{\infty} e^{- \Lambda \sqrt{z^4 +Az^2 +B}}\mathrm{d}z$,

with $A = 2(s^2-x^2)$ and $B = s^4 + x^4 + 2s^2x^2$.

Any help with a solution would be much appreciated.

Edit: I forgot to mention that the $s = kL$ where $L$ is a fixed value and I will eventually take a limit in which $k \rightarrow 0$, so there are opportunities for series expansions. I have tried the obvious by expanding the square root in powers of $k$, but there are then convergence issues in the region $|z-x| < k$.

A closed form solution is looking less and less likely as I try all the tricks I know and scour Gradshteyn, so a first term in $k$ (Edit: I originally said in $a$, that was a mistake) would also be much appreciated.

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The case $\Lambda = 1$, $a = 0$, $b = 1$ might be complicated enough. – Shai Covo Mar 22 '11 at 23:14
Just to be clear, is the $a = s^2 - x^2$ in the first equation different from the $a = s/L$ that will approach zero? – mjqxxxx Mar 23 '11 at 2:04
Sorry my bad! They are different $a$'s I will edit that in the question. – Kyle Mar 23 '11 at 2:31
Are you interested in the behavior for small $s$, or actually at $s=0$? – mjqxxxx Mar 23 '11 at 4:40
There's no match in the Inverse Symbolic Calculator for the numerical value $\int_0^\infty \exp(-\sqrt{x^4+1}) dx \approx 0.443587383072818$. – Hans Lundmark Mar 23 '11 at 8:06

Would a double expansion, once for the square root and once for the exponential, work? It seems like you're interested in the $b\rightarrow 0$ behavior so I tried it and the answer is of the form: $I=\sqrt{\frac{\pi}{4\Lambda}}e^{-\Lambda a}-\frac{b^{2}\pi}{4\sqrt{a}} erf(\Lambda a)+ O(b^{6})$ (The 4th order cancels). Cheers.

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(too long for a comment)

I am rather sure that the integral as it stands has no closed form in itself; however, one might be able to derive a series with elliptic integral terms (you do have the square root of a quartic, after all) that hopefully quickly converges (and computing an elliptic integral is quite easy with the AGM).

I also note that if you express your quartic's coefficients wholly in terms of $s$ and $x$, your quartic factors into two quadratics with complex conjugate roots (due to the constraints you put on those two parameters):

$$(z^2+a)^2+b^2=(z^2-2xz+s^2+x^2)(z^2+2xz+s^2+x^2)$$

and letting $\mu=x+is$, the quartic can also be expressed as $(z^2-2\Re\mu z+|\mu|^2)(z^2+2\Re\mu z+|\mu|^2)=(z-\mu)(z-\bar{\mu})(z+\mu)(z+\bar{\mu})$, such that the only parameters you have to contend with in your integral are $\mu$ and $\Lambda$.

I'll edit this once I figure out how to derive the eliiptic integral series...

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Hey! Welcome back :-) – Aryabhata Apr 7 '11 at 2:41
Hi Mo, nice to see you. :) – J. M. Apr 7 '11 at 2:42

When $a=0$ , $b$ is a real number,

Then $\int_0^\infty e^{-\Lambda\sqrt{(z^2+a)^2+b^2}}~dz$

$=\int_0^\infty e^{-\Lambda\sqrt{z^4+b^2}}~dz$

$=\int_0^\infty e^{-\Lambda\sqrt{(\sqrt{|b|\sinh z})^4+b^2}}~d(\sqrt{|b|\sinh z})$

$=\dfrac{\sqrt{|b|}}{2}\int_0^\infty\dfrac{e^{-\Lambda\sqrt{b^2\sinh^2z+b^2}}\cosh z}{\sqrt{\sinh z}}dz$

$=\dfrac{\sqrt{|b|}}{2}\int_0^\infty\dfrac{e^{-|b|\Lambda\cosh z}\cosh z}{\sqrt{\sinh z}}dz$

$=-\dfrac{\sqrt{|b|}}{2}\dfrac{d}{dx}\int_0^\infty\dfrac{e^{-x\cosh z}}{\sqrt{\sinh z}}dz~(x=|b|\Lambda)$

$=-\dfrac{\sqrt{|b|}}{2}\dfrac{d}{dx}\dfrac{\Gamma\left(\dfrac{1}{4}\right)\sqrt[4]x~K_{-\frac{1}{4}}(x)}{\sqrt[4]2\sqrt\pi}(x=|b|\Lambda)$ (according to http://people.math.sfu.ca/~cbm/aands/page_376.htm)

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