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In a course, my teacher told us that the following integral is convergent and used the comparison test to prove it; my question is how to find the antiderivative in closed form? It seems to exist; if, however, it doesn't exist, can someone prove it?

$$\int\sqrt{\dfrac1{1+x^3}}\mathrm dx$$

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What do you mean by "exists"? To clarify, the antiderivative is a function, you can compute its values at various points (so long as you specify $C$), but not a so-called "elementary" function (the kind of functions you are probably used to). –  Alex Becker Apr 17 '11 at 4:53
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If anybody's interested, I'll write up the way to derive the solution in terms of elliptic integrals. Mathematica's results are damned messy... –  J. M. Apr 17 '11 at 5:02
    
@J.M.- I will be more than interested to see this integral in action in an analytical method of solving. Thanks. –  night owl Apr 17 '11 at 5:16
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I'll give the details later... but here's the analytical solution: $$\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1\right)‌​\mid\frac{2+\sqrt{3}}{4}\right)$$ –  J. M. Apr 17 '11 at 5:36
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@awllower: The final result is expressed in terms of "elliptic integrals", not "elliptic functions"; however, one can use elliptic functions to derive the elliptic integral result (which is what I did). –  J. M. Apr 17 '11 at 16:19

1 Answer 1

up vote 15 down vote accepted

The first thing to do is to note that

$$x^3+1=(x+1)(x^2-x+1)$$

(one real and two complex conjugate roots). Using Jacobian elliptic functions requires having a quartic within the square root (the alternative of using Weierstrass elliptic functions is fine with square roots of cubics, but I'll leave that approach to someone else); the good thing is that by choosing a proper Möbius transformation, one can turn a cubic into a quartic (the algebraic geometers here might want to say a bit more than I have).

For the integral in question, the Möbius substitution needed is $x=\frac{-1+\sqrt{(-1)^2-(-1)+1}+(-1-\sqrt{(-1)^2-(-1)+1})v}{1+v}=\frac{2\sqrt{3}}{1+v}-(1+\sqrt{3})$; we then have

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=-2\int\frac{\mathrm dv}{\sqrt{(1-v^2)(2\sqrt{3}-3+(2\sqrt{3}+3)v^2)}}$$

At this point, making use of the Jacobian elliptic function identity $\mathrm{sn}^2(u|m)+\mathrm{cn}^2(u|m)=1$ (nothing more than the usual Pythagorean identity in elliptic function garb), we could make either of the substitutions $v=\mathrm{sn}(u|m)$ or $v=\mathrm{cn}(u|m)$. The latter is a bit more convenient, since $\mathrm dv=-\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du$, which can conveniently get rid of the negative sign in the integral. Thus, the integral turns into

$$2\int\frac{\mathrm{sn}(u|m)\mathrm{dn}(u|m)\mathrm du}{\sqrt{(1-\mathrm{cn}^2(u|m))(2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m))}}$$

or (by using the Pythagorean identity)

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}$$

Here, one now chooses a proper value of $m$ such that the integrand reduces to a constant. Skipping the details, we let $m=\frac{2+\sqrt{3}}{4}$ such that

$$2\int\frac{\mathrm{dn}(u|m)\mathrm du}{\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(u|m)}}=\int\frac{\mathrm du}{\sqrt[4]{3}}$$

To undo the substitutions, we note that $u=F(\arccos(v)|m)$ and $v=\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1$, giving the final result

$$\int\frac{\mathrm dx}{\sqrt{x^3+1}}=\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+x}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$

This result can be verified by differentiating the right hand side (remember that $\frac{\mathrm d}{\mathrm d\phi}F(\phi|m)=\frac1{\sqrt{1-m\sin^2\phi}}$) and noting that it is the same as the integrand.

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Could you recommend a book containing the special functions and integration techniques you're using? These things have always been sort of a mystery to me. –  t.b. Apr 18 '11 at 9:45
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@Theo: at least for the elliptic integrals, I got my bag of tricks from Byrd/Friedman; I have to admit I'm still actively learning the tricks myself since one of my recent projects involves a great deal of elliptic integral manipulations. –  J. M. Apr 18 '11 at 9:48
    
Also, I have been told that it might be more profitable to do manipulations with the Carlson symmetric elliptic integrals instead, but I have yet to study the requisite papers by Carlson. –  J. M. Apr 18 '11 at 9:52
    
Thanks a lot. I'll check this book out, the big G doesn't let me have a peek, so I've ordered it in the library. Good luck with your project! (Ah, I didn't see your second comment before posting, thanks for that, too.) –  t.b. Apr 18 '11 at 9:55
    
Thanks a lot, @J.M.It really helps me a lot. –  awllower Apr 19 '11 at 8:19

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