What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$

I came across this integration in a physics problem. I suspect role of complex numbers here. '$a$' is a constant

  • $\begingroup$ The primitive function cannot be expressed in terms of 'normal functions'. You might be able to do it in some general cases, like for $a=0$ the integral is $-\frac{2x}{\sqrt{x^3}}$. $\endgroup$ – Botond Mar 10 '18 at 16:41
  • $\begingroup$ Complex analysis will hardly help here unless this was a Riemann integral: with limits and stuff...and even then it is not sure. $\endgroup$ – DonAntonio Mar 10 '18 at 16:45
  • $\begingroup$ a is not equal to zero it has physical value $\endgroup$ – 3.14159 Mar 10 '18 at 16:47
  • $\begingroup$ @3.14159 What kind of physics problem do you need it for? You might need to use approximations or a different method. $\endgroup$ – Botond Mar 10 '18 at 17:01
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    $\begingroup$ @3.14159 You are free to add it as the context of your integral. Or you can just write it down here. $\endgroup$ – Botond Mar 10 '18 at 17:26

As the other users said, this integral is unlikely to have an elementary form, however, it can be expressed in terms of the well known Gauss hypergeometric function, which can be easily evaluated by most CAS or even Wolfram Alpha.

First, let's consider the case $|x|<|a|$, then we can substitute:

$$x=at, \qquad |t|<1$$

$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{dt}{\sqrt{1+t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{3k} dt=$$

$$=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{3k+1}}{3k+1}=a^{-1/2} \Gamma \left( \frac{1}{2} \right) \sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k+1}}{3k+1}=$$

$$=\frac{1}{3}\sqrt{\frac{\pi}{a}}~t~\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}$$

To find the hypergeometric form of the series above, we consider the ratio of the successive terms:

$$\frac{c_{k+1}}{c_k}=\frac{\left(-1/2-k \right)(k+1/3)}{ (k+4/3)} \frac{t^3}{k+1}=\frac{\left(k+1/2 \right)(k+1/3)}{ (k+4/3)} \frac{-t^3}{k+1}$$


Which, by definition makes the series:

$$\sum_{k=0}^\infty \frac{1}{\Gamma \left(1/2-k \right) k!} \frac{t^{3k}}{k+1/3}=\frac{3}{\sqrt{\pi}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)$$

Which makes the integral:

$$\int\frac{dx}{\sqrt{x^3+a^3}}=\frac{t}{\sqrt{a}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -t^3 \right)=\frac{x}{\sqrt{a^3}} {_2F_1} \left( \frac{1}{2}, \frac{1}{3}; \frac{4}{3}; -\frac{x^3}{a^3} \right)$$

This is a correct answer for $|x|<|a|$, as can be checked by numerical experiments.

Mathematica, or other advanced software, can directly evaluate and plot hypergeometric function, which makes this form more useful than the original integral.

For $|x|>|a|$ we can use the same method of binomial expansion to get the hypergeometric form.

$$x=at, \qquad |t|>1$$

$$\int\frac{dx}{\sqrt{x^3+a^3}}=a^{-1/2}\int\frac{t^{-3/2} dt}{\sqrt{1+1/t^3}}=a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \int t^{-3k-3/2} dt=$$

$$=-a^{-1/2} \sum_{k=0}^\infty \binom{-1/2}{k} \frac{t^{-3k-1/2}}{3k+1/2}$$

It's straightforward to continue in the same way and obtain another hypergeometric function.

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It's this:

$$\frac{2 \sqrt[6]{-1} \sqrt[3]{a^3} \sqrt{(-1)^{5/6} \left(\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}-1\right)} \sqrt{\frac{(-1)^{2/3} x^2}{\left(a^3\right)^{2/3}}+\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}+1} F\left(\left.\sin ^{-1}\left(\frac{\sqrt{-\frac{(-1)^{5/6} x}{\sqrt[3]{a^3}}-(-1)^{5/6}}}{\sqrt[4]{3}}\right)\right|\sqrt[3]{-1}\right)}{\sqrt[4]{3} \sqrt{a^3+x^3}}$$ where $F$ denotes the Elliptic Integral.

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  • $\begingroup$ What is $F$, what is $\sin^{-1}$? $\endgroup$ – klirk Mar 10 '18 at 17:00
  • $\begingroup$ @klirk $\sin^{-1}$ is an ugly way to denote the arcsine function, whilst $F$ denotes the Elliptic Integral. This integral cannot be solved in terms of elementary functions, and Elliptic Functions are a special type of functions you will surely meet many times in the future! $\endgroup$ – Mycroft Mar 10 '18 at 17:02
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    $\begingroup$ The formatting could really use some work $\endgroup$ – Yuriy S Mar 10 '18 at 17:02
  • $\begingroup$ @VonNeumann: Why don't you include this information in your answer? $\endgroup$ – klirk Mar 10 '18 at 17:06
  • $\begingroup$ @klirk Because they are supposed to be very elementary questions. If I write $\pi$ without specifying something, you won't surely think about a strange mystery. $\endgroup$ – Mycroft Mar 10 '18 at 17:10

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