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I'm stumped. How does one solve the following expression: $$\int\left(\frac{x^4}{1-x^4}\right)^kdx, k\in\mathbb{N}$$ I thought it would be a fun challenge, but it beats me.

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    $\begingroup$ Use partial fractions? $$\dfrac{x^4}{1-x^4} = \dfrac{1}{2(1+x^2)}+\dfrac{1}{4(1+x)}+\dfrac{1}{4(1-x)}-1$$ $\endgroup$ Commented Jan 29, 2020 at 18:29

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$$ I_n = \int\left(\frac{x^4}{1-x^4}\right)^n dx . $$

This is a rational function, so there is an answer in terms of rational functions, logarithms, and arctangents. Depedence on $n$ is complicated.

We can get a formula in terms of $n$ like this, however: $$ I_n = {\frac {1}{4\,n+1} \left( {\frac {{x}^{4}}{1-{x}^{4}}} \right) ^{n+1/4} {\mbox{$_2$F$_1$}\left(\frac54,n+\frac14;\,n+\frac54;\,-{\frac {{x}^{4}}{1-{x}^{4}}}\right)} } $$ This should hold even if $n$ is not an integer.


explanation
Change variables $y = \frac{x^4}{1-x^4}$ to get $$ I_n = \int\frac{y^{n-3/4}}{4(y+1)^{5/4}}\;dy $$ Next, note that $$ \frac{y^{n-3/4}}{4(y+1)^{5/4}} = \frac{y^{n-3/4}}{4}\;{}_1F_0\left(\frac54;;-y\right) = \frac{y^{n-3/4}}{4}\;{}_2F_1\left(\frac54,n+\frac14;n+\frac14;-y\right) $$ Now we use a hypergeometric differentiation formula, Gradshteyn & Ryzhik 9.103.3

Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.

$$ \frac{d}{dy}\left[y^{c-1}\;{}_2F_1\left(a,b;c;-y\right)\right]= (c-1)y^{c-2}\;{}_2F_1\left(a,b;c-1;-y\right) \tag{9.103.3}$$

Combining these yields the result $$ I_n = {\frac {1}{4\,n+1} \left( {\frac {{x}^{4}}{1-{x}^{4}}} \right) ^{n+1/4} {\mbox{$_2$F$_1$}\left(\frac54,n+\frac14;\,n+\frac54;\,-{\frac {{x}^{4}}{1-{x}^{4}}}\right)} } $$

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You can obtain a recurrence relation and then find a formula using sums and products for a closed form solution. $k\geq 1$, let $$ I_k(x) = \int \left( \frac{x^4}{1-x^4} \right) ^k {\rm d}x$$

Using by parts with $u = (1-x^4)^{-k}$ and ${\rm d}v = x^{4k} {\rm d}x$, we will get

$$ \begin {align} I_k(x) &= \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} \int \frac{x^{4(k+1)}}{(1-x^4)^{k+1}} { \rm d}x \\ & = \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} I_{k+1}(x) \end {align} $$

Rewriting, we get

$$ I_{k+1}(x) = \frac{x^{4k+1}}{4k(1-x^4)^k} - \frac{4k+1}{4k} I_k(x) $$

Now, getting a formula for $I_k$ in terms of sums might be a little tedious but shouldn't be too difficult. First solve it for $k=1$ using partial fractions, $$ I_1(x) = \frac12 \arctan(x) +\frac14 \ln \left| \frac{1+x}{1-x} \right| -x + C$$

Then, for $k \geq 2$,

$$ I_k(x) = I_1(x) \prod_{n=1}^{k-1} \left(-\frac{4n+1}{4n} \right) + \sum_{n=1}^{k-1} \left( \prod_{m=n+1}^{k-1} -\frac{4m+1}{4m} \right)\frac{x^{4n+1}}{4n(1-x^4)^n} +C $$

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