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Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$

$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as

$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$

So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$

So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$

Now How can I solve it, Help me

Thanks

graydad
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juantheron
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  • The final answer seems to be $\frac{\pi}{5}\sqrt{1+2/\sqrt{5}}$... – mickep Sep 20 '15 at 16:57
  • You have $$\displaystyle\int\limits_0^1\sum\limits_{n=0}^\infty\Big(x^{5n}-x^{3+5n}\Big) ,\mathrm{d}x,$$ so it would just be $$\displaystyle \sum\limits_{n=0}^\infty\int\limits_0^1\Big(x^{5n}-x^{3+5n}\Big) ,\mathrm{d}x = \sum\limits_{n=0}^\infty\left(\frac{3}{(1+5n)(4+5n)}\right) = {\frac{\pi}{5}}\sqrt{1+\frac{2}{\sqrt{5}}},$$ wouldn't it? – dbanet Sep 20 '15 at 17:16

4 Answers4

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Hint: $$ \begin{align}\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx&= 1-{1\over4}+{1\over6}-{1\over9}+{1\over11}-{1\over14}+\ldots\\&= \sum_{k=0}^\infty{3\over(5k+1)(5k+4)}\end{align} $$

Intelligenti pauca
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3

To expand the answer posted by @Arentino, we have

$$\begin{align} S&=3\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+4)}\\\\ &=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k+4/5}\right)\\\\ &=\frac34+\frac15\sum_{k=1}^\infty\left(\frac{1}{k+1/5}-\frac{1}{k}\right)+\frac15\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+4/5}\right)\\\\ &=\frac34+\frac15(\psi(9/5)-\psi(6/5)) \tag 1 \end{align}$$

Using the reflection and recurrence formulae of the digamma function in $(1)$ reveals that

$$\begin{align} S&=\frac34+\frac15(\psi(9/5)-\psi(6/5))\\\\ &=\frac34+\frac15\left(\psi(4/5)+\frac54\right)-\frac15\left(\psi(1/5)+5\right)\\\\ &=\frac34+\frac15\left(\psi(4/5)-\psi(1/5)\right)+\left(\frac14-1\right)\\\\ &=\frac15 \pi \cot(\pi/5)\\\\\ &=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}} \end{align}$$

as expected!!

Mark Viola
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  • (+1) Well done, but that can be shortened a little by recalling that $$\sum_{k\geq 0}\left(\frac{1}{k+a}-\frac{1}{k+b}\right) = \psi(b)-\psi(a).$$ – Jack D'Aurizio Sep 20 '15 at 19:41
  • @JackD'Aurizio Thanks! I suppose I derived that "shortened result" from the reflection and recurrence identities. – Mark Viola Sep 20 '15 at 19:50
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Here is a sketch of a way that does not use the digamma function.

If we start with your own favourite(?) change of variable, $$ u=\frac{1}{x}-x, $$ then, we end up with $$ \int_0^{+\infty}\frac{2+u^2}{5+5u^2+u^4}\,du-\int_0^{+\infty}\frac{1}{\sqrt{4+u^2}(5+5u^2+u^4)}\,du $$ In the first integral, it is easier (than in the original one) to do partial fraction decomposition. In the second integral, we let $$ t=\frac{u}{\sqrt{4+u^2}}, $$ which will give us the integral $$ \int_0^1\frac{1-t^2}{5+10t^2+t^4}\,dt, $$ which also can be done using partial fraction decomposition (don't forget that it should come with a minus). I leave the funny calculations to you...

The final result (as already mentioned in several places before) turns out to be $$ \frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}. $$ Probably there are smarter substitutions, leading to an even simpler expression to integrate.

mickep
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$$\begin{eqnarray*}\sum_{k=0}^{+\infty}\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right)&=&\int_{0}^{1}\frac{1-x^3}{1-x^5}\,dx\\&=&\int_{0}^{1}\frac{1+\frac{1}{\sqrt{5}}}{2+(1-\sqrt{5})x+2x^2}\,dx+\int_{0}^{1}\frac{1-\frac{1}{\sqrt{5}}}{2+(1+\sqrt{5})x+2x^2}\,dx\end{eqnarray*} $$ where the last identity comes from the residue theorem.

By using the reflection formula for the $\psi$ function, we have that the original series equals: $$ \frac{1}{5}\left(\psi\left(\frac{4}{5}\right)-\psi\left(\frac{1}{5}\right)\right) =\frac{\pi}{5}\cot\frac{\pi}{5}.$$

Jack D'Aurizio
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