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.

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?

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

share|improve this question
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
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
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
@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 16 down vote accepted

The first thing to do is to note that


(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.

share|improve this answer
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
@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

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.