Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

(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):


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...

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.