7
$\begingroup$

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.

  • 0
    http://www.wolframalpha.com/input/?i=Integrate+1%2Fsqrt%283x%5E4%2B6x%5E2-1%29+from+1+to+infinity gives a result with two elliptic functions, $i$ and $\sinh^{-1}$. It will also do the indefinite integral.2011-04-04

1 Answers 1

11

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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    Thanks, JM. I agree about Mathematica. Maple, as well.2011-04-07