<s>[INST] Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $n = 1$, the integrand $=x+x^3$
For $n = 2$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}$
For $n = 3$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}$
and so on.
For $n = 1000$, the integrand $=x+x^3-x^5-x^7+x^9+x^{11}-x^{13}-x^{15}+x^{17}+x^{19}-\dots+x^{7993}+x^{7995}$
Using MS-EXCEL with $n=1000$, I found that the value is approximately $0.56...$.
I do not know if  $n \rightarrow \infty$ , will the required expression have a closed form or no.

Your help would be appreciated. THANKS!
 [/INST] We seek to calculate $$\begin{align}\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx&= \frac{1}{2}+\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{10}+\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\cdots \\&=\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots\right)\right)\end{align}$$ Now $$\arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots \implies \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots=\arctan{1}=\frac{\pi}{4}$$ and $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \implies \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)=\frac{\ln2}{2}$$ Hence our answer is $\frac{\pi}{8}+\frac{\ln2}{4}$.
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