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

$$\int_{1}^{\infty}\frac{1}{\sqrt{3x^{4}+6x^{2}-1}}dx$$

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.

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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
    
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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
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1 Answer

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

$$\frac1{\sqrt{u+v}}F\left(\arcsin\left(\sqrt{\frac{u+v}{v+uv}}\right)\mid\frac{v}{u+v}\right)$$

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

$$\frac1{\sqrt[4]{48}}F\left(\arcsin\left(\sqrt{\sqrt{3}-1}\right)\mid\frac{2+\sqrt{3}}{4}\right)$$

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...

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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
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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
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