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I need to compute: $$\int_0^1 \frac{1-x^3}{1-x^5} dx$$

I tried integrating it by partial fractions but couldn't succeed. Is there any other way to integrate this?

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    [W|A](http://tinyurl.com/bpg8jb4) gives a very interesting indefinite integral, which I can't explain. Plugin value and you [get](http://tinyurl.com/d6wgj62): $0.864806$.2012-06-26
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    Probably, "the shortest path between two statements in the real world is through the complex world" :-)2012-06-26
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    @draks: If $f$ and $g$ have no common roots and $g(x)$ has no double roots, residue calculus gives $$\frac{f(x)} {g(x)}=\sum_{g(r)=0}\frac{f(r)/g\,'(r)}{x-r}.$$ This is obtained by applying contour integration and the residue theorem. The sum is taken over all roots $r$ of the $g$. Therefore by discarding the common root via geometric sum formula, we have $$\frac{(x^3-1)/(x-1)}{(x^5-1)/(x-1)}=\frac{x^2+x+1}{x^4+x^3+x^2+x+1} \\ =\sum_{\substack{\omega^5=1 \\ \omega\ne1}}\frac{\omega^2+\omega+1}{4\omega^3+3\omega^2+2\omega+1}\frac{1}{x- \omega}.$$ This should explain the indefinite integral.2012-06-26
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    @anon +1 nice $ $2012-06-26

3 Answers 3

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Use partial fraction:

$$ \frac{1-x^3}{1-x^5} = \frac{\sqrt{5}-1}{\sqrt{5}(2x^2+(1+\sqrt{5})x+2)}+\frac{\sqrt{5} +1}{\sqrt{5}(2x^2+(1-\sqrt{5})x+2)} $$

and then integrate both summands with

$$ \int 1/(ax^2+bx+c)dx= 2\frac{\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}{\left(\sqrt{4ac-b^2}\right)} + \text{constant}, $$ which is done by completing the square and scaling/translating the denominator to the form $A^2y^2+C^2$. You'll get $\int 1 /(A^2y^2+C^2)dy = \frac{\tan^{-1}(Ay/C)}{AC}+\text{constant}$.

Plugin your limits and you're done. I leave the hard substitution work to you. Ship ahoi.

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    (For fun, also note the golden ratio.) You might mention that the integration you have there is done by completing the square and scaling/translating.2012-06-26
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    ;-) Thanks for pointing this out, click on "show step" at W|a.2012-06-26
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    Thanks Sir @draks . Btw, no other way except partial fraction?2012-06-26
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    @Bazinga, I sailed around the whole world, but found no other, yet...;-)2012-06-26
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An alternative approach would be to use the geometric series $$ \frac{1-x^3}{1-x^5} = \sum_{n=0}^{\infty} (x^{5n} - x^{5n+3}) $$ to evaluate the integral as $$\int_0^1 \frac{1-x^3}{1-x^5} dx = \sum_{n=0}^{\infty} \Big(\frac{1}{5n+1} - \frac{1}{5n+4} \Big)$$ Using the digamma function this can now be evaluated as $$ \sum_{n=0}^{\infty} \Big(\frac{1}{5n+1} - \frac{1}{5n+4} \Big) = \frac{1}{5} (\psi(4/5) - \psi(1/5) ) $$ In conclusion one can note that the digamma function of a rational number in the interal $]0;1[$ can always be expressed as a sum of elementary functions (trigonometric functions and logarithms of trigonometric functions) evaluated at a rational number times $\pi$.

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    Amusingly, the OP is using the integral to try to evaluate that sum, as you can see here: http://math.stackexchange.com/questions/163165/how-to-find-the-sum-of-this-infinite-series2012-06-26
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Another method, using the Beta function. Let $t=x^5$, then $$ \begin{align} \int_0^1 \frac{1-x^3}{1-x^5} dx &=\lim_{\delta\to0}\frac15\int_0^1(t^{-4/5}-t^{-1/5})(1-t)^{\delta-1}\,\mathrm{d}t\\ &=\lim_{\delta\to0}\frac15\left(\mathrm{B}(1/5,\delta)-\mathrm{B}(4/5,\delta)\right)\\ &=\lim_{\delta\to0}\frac15\left(\frac{\Gamma(1/5)\Gamma(\delta)}{\Gamma(1/5+\delta)}-\frac{\Gamma(4/5)\Gamma(\delta)}{\Gamma(4/5+\delta)}\right)\\ &=\lim_{\delta\to0}\frac15\left(\frac{\Gamma(1/5)}{\Gamma(1/5+\delta)}-\frac{\Gamma(4/5)}{\Gamma(4/5+\delta)}\right)\frac{\Gamma(1+\delta)}{\delta}\\ &=\frac15\left(\frac{\Gamma'(4/5)}{\Gamma(4/5)}-\frac{\Gamma'(1/5)}{\Gamma(1/5)}\right)\\ &=\frac15(\psi(4/5)-\psi(1/5))\\ &=\frac{\pi}{5}\cot\left(\frac{\pi}{5}\right)\\ &=\frac{\pi}{5}\sqrt{\frac{5+2\sqrt{5}}{5}} \end{align} $$ Using the identity $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$