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Let $I_n=\int_{0}^{1} \frac{x^n}{\sqrt{x^3+1}}dx$. Show that $(2n−1)I_n+2(n−2)I_{n-3}= 2\sqrt{2}$ for all $n\geq3$. Then compute $I_8$.

How do I compute that integral? My idea was to reduce it to some known forms, as $\int\frac{1}{\sqrt{1-x^2}}dx$... Is there any way to solve the problem without computing the integral though?

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  • $\begingroup$ Please verify the edit I made is accurate. $\endgroup$ – tilper Nov 8 '16 at 15:29
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    $\begingroup$ I imagine one of the indices should be $n-1$ or $n+1$... $\endgroup$ – TastyRomeo Nov 8 '16 at 15:32
  • $\begingroup$ Sorry! This is the correct edit $\endgroup$ – user368484 Nov 8 '16 at 15:36
  • $\begingroup$ I would start with integration by parts $u = x^{n-2}, dv = \frac {x^2}{\sqrt {1-x^3}} dx$ This should then give you $I_n$ in terms of $I_{n-1}$ and $I_{n-2}$ $\endgroup$ – Doug M Nov 8 '16 at 15:37
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I doubt that it is a good idea to use the hypergeometric function. Also substitution is not working. Actually it is pretty easy to use integration by parts to get the answer. Define $$ J_n=(2n−1)I_n+2(n−2)I_{n-3}, n\ge 3.$$ Note \begin{eqnarray} J_n&=&(2n−1)I_n+2(n−2)I_{n-3}\\ &=&\int_0^{1}\frac{(2n-1)x^n+(2n-4)x^{n-3}}{\sqrt{x^3+1}}dx\\ &=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)x^3+(2n-4)\bigg]}{\sqrt{x^3+1}}dx\\ &=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)(x^3+1)-3\bigg]}{\sqrt{x^3+1}}dx\\ &=&(2n-1)\int_0^{1}x^{n-3}\sqrt{x^3+1}dx-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &=&\frac{2n-1}{n-2}\int_0^{1}\sqrt{x^3+1}d(x^{n-2})-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &=&\frac{2n-1}{n-2}\left(\sqrt{x^3+1}x^{n-2}\bigg|_0^1-\frac{3}{2}\int_0^1\frac{x^{n}}{\sqrt{x^3+1}}dx\right)-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &=&\frac{2n-1}{n-2}\sqrt2-\frac{3}{2n-4}J_n\\ \end{eqnarray} and hence $$ \left(1+\frac{3}{2n-4}\right)J_n=\frac{2n-1}{n-2}\sqrt2$$ which gives $J_n=2\sqrt2$.

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  • $\begingroup$ +1 from me. I just had fun in solving the integral, not in finding a recurrence :) $\endgroup$ – Von Neumann Nov 9 '16 at 15:02
  • $\begingroup$ And this one is from Mr. Trump! :-) $\endgroup$ – mrs Nov 9 '16 at 15:39
  • $\begingroup$ @BabakS., congrats, Mr. Trump! $\endgroup$ – xpaul Nov 9 '16 at 15:40
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Just for the sake of amusement in solving integrals.

You can always mind in this way. The integration is bounded between $0$ and $1$ hence you may make use of the binomial series:

$$\frac{1}{\sqrt{x^3+1}} = (1+x^3)^{-1/2} = \sum_{k = 0}^{+\infty} \binom{-1/2}{k} x^{3k}$$

Hence

$$\int_0^1 \frac{x^n}{\sqrt{1+x^3}}\ \text{d}x = \sum_{k = 0}^{+\infty} \binom{-1/2}{k}\int_0^1 x^{3k + n}\ \text{d}x = \sum_{k = 0}^{+\infty} \binom{-1/2}{k} \frac{1}{3k + n + 1}$$

The last series does converge and its sum is expressed in terms of HyperGeometric Function:

$$\sum_{k = 0}^{+\infty} \binom{-1/2}{k} \frac{1}{3k + n + 1} = \frac{\, _2F_1\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right)}{n+1}$$

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    $\begingroup$ +1 How about using diferential binomial? $\endgroup$ – mrs Nov 8 '16 at 16:10
  • $\begingroup$ @Babak, see my answer. $\endgroup$ – xpaul Nov 9 '16 at 15:01
  • $\begingroup$ Maybe it is a stupid question but normally,in order to express this term in binominal series, should not we have a complex number instead of $\frac{1}{2}$ ? $\endgroup$ – optimal control Nov 18 '16 at 20:19
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HINT

Notice that if you let $u = x^3+1$ then $du = 3x^2dx$ so $$ \int \frac{x^2 dx}{\sqrt{x^3+1}} = \frac{1}{3} \int u^{-1/2}du $$ and apply integration by parts to your original integral

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  • $\begingroup$ Easy peasy!! :D +1 $\endgroup$ – Von Neumann Nov 8 '16 at 15:37
  • $\begingroup$ @AlanTuring :-) $\endgroup$ – gt6989b Nov 8 '16 at 15:38
  • $\begingroup$ Right, that should give the solution quite neatly $\endgroup$ – user368484 Nov 8 '16 at 15:39
  • $\begingroup$ Sorry, but I'm still a newbie with integrals :) $\endgroup$ – user368484 Nov 8 '16 at 15:41
  • $\begingroup$ @0inegue nothing to be sorry about :), just learn the technique... taking integrals, unlike derivatives (which are mechanical), is really an artform... $\endgroup$ – gt6989b Nov 8 '16 at 15:44
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                    **This is another great way to compute $I_8$.**

As we have $I_8=\int_0^1\frac{x^8}{\sqrt{1+x^3}}dx$. Let $1+x^3=t^2$. You may ask me why am I doing this? I do it because we can do that by binomial_differentials. So you can overcome the indefinite integral first and the do for interval $[0,1]$.

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This is a partial solution. There is a problem at the end. If you can resolve it, please do so in a comment and I will fix my solution, or post it as a separate solution.

Here is a correct solution that avoids the problem of going from the incomplete beta function to Gauss's hypergeometric function.

Let $y=x^{3}$ \begin{align} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx &= \frac{1}{3} \int\limits_{0}^{1} \frac{y^{(n-2)/3}}{\sqrt{y+1}} dy \\ &= \frac{1}{3} \frac{\Gamma(\frac{n+1}{3})\Gamma(1)}{\Gamma(\frac{n+4}{3})}\,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right) \\ &= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right) \end{align}

We used the analytic continuation of Gauss's hypergeometric function \begin{equation} {}_{2}\mathrm{F}_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c - b)} \int\limits_{0}^{1} t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} dt \end{equation} For $\mathrm{Re}\, c \gt \mathrm{Re}\, b \gt 0$, $|\mathrm{arg}(1-z)| \lt \pi$

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  • $\begingroup$ +1 because of the cool use of the Incomplete Beta function! :D $\endgroup$ – Von Neumann Nov 8 '16 at 16:44

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