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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.

  • 2
    The case $\Lambda = 1$, $a = 0$, $b = 1$ might be complicated enough.2011-03-22
  • 0
    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?2011-03-23
  • 0
    Sorry my bad! They are different $a$'s I will edit that in the question.2011-03-23
  • 0
    Are you interested in the behavior for small $s$, or actually at $s=0$?2011-03-23
  • 0
    I am interested in the behavior for small s.2011-03-23
  • 2
    There's no match in the Inverse Symbolic Calculator for the numerical value $\int_0^\infty \exp(-\sqrt{x^4+1}) dx \approx 0.443587383072818$.2011-03-23
  • 0
    What does $\Lambda$ represent? Is it arbitrary or are you only interested in large values of $\Lambda$? That may also help.2011-03-23

3 Answers 3