Take the 2-minute tour ×
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.

Here is a rather tough integration. I think it looks like an Elliptic Integral of some sort.


Since there are no odd terms in the quartic, I thought maybe completing the square

would be OK. But I got nowhere. I even factored it into:

$3x^{4}+6x^2-1=((2\sqrt{3}-3)x^{2}+1)((2\sqrt{3}+3)x^{2}-1)$ and got nowhere.

I think the solution will involve the Gamma function in some manner.

Does anyone have a good starting point for this... like a clever substitution?.

Thanks for any input.

share|improve this question
What is the source of the problem? What kind of answer are you looking for? Numeric approximation? "Closed" form? –  Aryabhata Mar 6 '11 at 21:33
Check out the page: everything2.com/title/elliptic+integral+standard+forms –  Fabian Mar 6 '11 at 22:50
Thank you all. I was looking for a closed form. I ran it through my Voyage 200 and got the .47369 as well. I was just wondering how it could be done by hand..if at all. It ends up being a difficult Elliptic, then I will forget about it. –  Cody Mar 7 '11 at 15:34
wolframalpha.com/input/… gives a result with two elliptic functions, $i$ and $\sinh^{-1}$. It will also do the indefinite integral. –  Ross Millikan Apr 4 '11 at 20:58
add comment

1 Answer 1

up vote 9 down vote accepted

As has been mentioned many times on this site, anytime you have an algebraic function containing the square root of a cubic or a quartic, you are bound to bump into an elliptic integral.

Usually, such things are handled by using Jacobian elliptic functions for substitutions (in a manner similar to using substitution with trigonometric or hyperbolic functions when you have the square root of a quadratic in an integral).

I'll skip the tedious details of figuring out the proper substitution, since Byrd and Friedman give a formula for handling your integral (formula 212.00 in their handbook):

$$\int_y^\infty\frac{\mathrm dt}{\sqrt{(t^2+a^2)(t^2-b^2)}}=\frac1{\sqrt{a^2+b^2}}F\left(\arcsin\left(\sqrt{\frac{a^2+b^2}{a^2+y^2}}\right) \mid\frac{a^2}{a^2+b^2}\right)$$

where $F(\phi|m)$ is the incomplete elliptic integral of the first kind.

Coming back to your integral, we let $u=2\sqrt{3}-3$ and $v=2\sqrt{3}+3$ such that

$$\int_1^\infty\frac{\mathrm dx}{\sqrt{3x^4+6x^2-1}}=\frac1{\sqrt{uv}}\int_1^\infty \frac{\mathrm dx}{\sqrt{(x^2+1/u)(x^2-1/v)}}$$

Using the quoted formula, the integral reduces to


Substituting the values of $u$ and $v$ into this expression and simplifying, we have the result


which agrees with the numerical result in the comments.

As an aside, I consider it a capital annoyance that Mathematica often returns results with complex amplitudes even for real results...

share|improve this answer
For those who want to puzzle out the needed substitution: $\mathrm{ns}^2(u|m)$ is involved in the required variable substitution. –  J. M. Apr 6 '11 at 9:26
M.: Welcome back! :) –  Mike Spivey Apr 6 '11 at 13:42
Thanks @Mike; won't be staying very long, but I'm exploiting a gap in time I have. –  J. M. Apr 7 '11 at 2:21
Thanks, JM. I agree about Mathematica. Maple, as well. –  Cody Apr 7 '11 at 10:35
add comment

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.