# Integral $\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$

$$\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$ Here's my attempts but I'm not sure that I'm doing well : $$\text{The integral gives :} 1-\frac1 4+\frac1 6-\frac1 9+ \frac{1}{11}-\frac{1}{14}+\dots$$ This series is : $$S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)}$$ Using Wolfram alpha I got : $$S=\frac1 5\sqrt{1+\frac{2}{\sqrt{5}}}\pi$$ So If what I did is true the integral must give us this value.

• Yes that's what I am asking about ! – user784351 May 6 at 2:32

Let : $$I=\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$ Let $$S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)}$$ Let's compute it : \begin{align} S&=3\sum_{k=0}^\infty \frac{1}{(5k+1)(5k+4)} \\\\ &=\frac3 4 +\frac1 5 \sum_{k=1}^\infty\bigg(\frac{1}{k+1/5}-\frac{1}{k+4/5}\bigg) \\\ &=\frac{3}{4}+\frac{1}5\sum_{k=1}^\infty\bigg(\frac{1}{k+1/5}-\frac1 k\bigg)+\frac1 5\sum_{k=1}^\infty\bigg(\frac1 k-\frac1{k+4/5}\bigg)\\\ \text{We are introducing the digamma function \psi}: \\\ S&=\frac 3 4+\frac 1 5 \Bigg(\psi\bigg(\frac{9}5\bigg)-\psi\bigg(\frac{6}5\bigg)\Bigg)\\\ &=\frac 3 4+\frac 1 5 \Bigg(\psi\bigg(\frac{4}5\bigg)+\frac{5}4\Bigg)-\frac1 5\Bigg(\psi\bigg(\frac1 5\bigg)+5\Bigg)\\\ &=\frac 3 4+\frac 1 5\Bigg(\psi\bigg(\frac4 5\bigg)-\psi\bigg(\frac1 5\bigg)\Bigg)+\bigg(\frac1 4-1\bigg)\\\ &=\frac{1}5 \pi \cot\bigg(\frac{\pi}{5}\bigg) \\\ &=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}} \end{align} Therefore : $$I=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}$$
Your series is $$\sum_{n=0}^{+\infty}(x^{5n}-x^{5n+3})=$$
$$(1-x^3)\sum_{n=0}^{+\infty}(x^5)^n$$