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.
 A: hint
Your series is $$\sum_{n=0}^{+\infty}(x^{5n}-x^{5n+3})=$$
$$(1-x^3)\sum_{n=0}^{+\infty}(x^5)^n$$
A: 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 :
$$I=\frac{\pi}{5}\sqrt{1+\frac{2}{\sqrt{5}}}$$
