I remember having to solve the following problem:

Define \begin{equation} I_n=\int_0^{1}\frac{x^ndx}{\sqrt{x^3+1}} \end{equation}

Prove: \begin{equation} \left(2n-1\right)I_n+2\left(n-2\right)I_{n-3}=2\sqrt{2} \end{equation}

And I found the solution. But in doing so, when trying different things, I decided to expand the integral using parts to find that

$$ I_n=\int_0^{1}\frac{x^ndx}{\sqrt{x^3+1}}\approx \bigg[x^n\cdot\int\frac{dx_1}{\sqrt{x^3+1}}-nx^{n-1}\iint\frac{dx_1 dx_2}{\sqrt{x^3+1}}$$ $$ +n\left(n-1\right)x^{n-2}\iiint\frac{dx_1 dx_2 dx_3}{\sqrt{x^3+1}}-n\left(n-1\right)\left(n-2\right)x^{n-3}\iiiint\frac{dx_1 dx_2 dx_3 dx_4}{\sqrt{x^3+1}}$$ \begin{equation} +\: ...\: +\left(-1\right)^{n}n!\underbrace{\int ... \int}_{n+1}\frac{dx_1\: ...\: dx_{n+1}}{\sqrt{x^3+1}}\bigg]_0^{1} \end{equation}

or something of the sort. So I became interested in elliptic integrals because I discovered that $\int\frac{dx}{\sqrt{x^3+1}}$ was not at all easy to solve, see here. But I wonder, what if I wanted to find the double, triple or quadruple integral of that thing as in my crude (and likely wrong) series above? Is it possible? So to sum up, is it possible to find

$$\iint\frac{dx_1 dx_2}{\sqrt{x^3+1}},\:\:\iiint\frac{dx_1 dx_2 dx_3}{\sqrt{x^3+1}}\:\:......$$

Also, a bit off topic, but are there any known methods for analyzing the integral series produced via integration by parts for those that don't repeat?

  • $\begingroup$ Should the $\sqrt{x^3-1}$ term (which is purely imaginary over $0<x<1$) at the top of your question be $\sqrt{x^3+1}$? $\endgroup$ – David H Oct 20 '14 at 13:52
  • $\begingroup$ You are right, I've fixed the error. $\endgroup$ – jm324354 Oct 20 '14 at 14:33
  • $\begingroup$ Your integrals at the end are written a bit strangely: Do you mean something like $$\int\int_0^{x_2}\frac{1}{\sqrt{x_1^3+1}}\,dx_1dx_2?$$ $\endgroup$ – Semiclassical Oct 20 '14 at 14:43
  • $\begingroup$ @Semiclassical Yes, but I do not wish to evaluate these integrals, simply keep them indefinite. What I meant by the $dx_1 dx_2$ is that I want to integrate with respect to $x$ twice. I added the sub-values to keep track of how many there were... $\endgroup$ – jm324354 Oct 20 '14 at 15:21
  • 1
    $\begingroup$ Right. It was more the lack of $x_1$ dependence in the integrals that was perplexing. As to the question itself, Cauchy's formula for repeated integration is where I'd try to start from. $\endgroup$ – Semiclassical Oct 20 '14 at 15:25

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