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 $n \in \mathbb{N}$

Question

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.

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Your help would be appreciated. THANKS!

Answer 1

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 $\frac{\pi}{8}+\frac{\ln2}{4}$.

Answer 2

Let

$$a = \int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$

Modified solution (now the main solution)

We have two choices: integral first then sum or vice versa.

Taking the integral first gives the following for the (finite) sum which we split immediately into even and odd terms

$$a=\sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} \frac{1}{2k}=\sum_{m=1}^{2n-1}(-1)^\frac{1+m}{2} \frac{1}{4m}+\sum_{m=1}^{2n-1}(-1)^\frac{1-m}{2} \frac{1}{4m-2}\tag{1}$$

Now we take the limit $n\to \infty$ to get the result $(4)$ of the original solution.

Claude Leibovici, in a comment, has taken the other altervative with the sum first.

Original solution

Exchanging integral and sum, and doing the limit in the sum gives

$$\sum_{k=1}^{\infty}(-1)^\frac{k^2+k+2}{2} \frac{1}{2k}$$

Now splitting even and odd summands

$k \to 2m$, $(-1)^{\frac{k^2+k+2}{2}}\to (-1)^{1+m+2m^2}\to (-1)^{1+m}$

$$s_{e} = \sum_{m=1}^{\infty} (-1)^{1+m} \frac{1}{4 m}=\frac{\log(2)}{4}\tag{2}$$

$k \to 2m-1$, $(-1)^{\frac{k^2+k+2}{2}}\to (-1)^{1-m+2m^2}\to (-1)^{1-m}$

$$s_{o} = \sum_{m=1}^{\infty} (-1)^{1+m} \frac{1}{4 m-2}=\frac{\pi}{8}\tag{3}$$

Hence we have

$$a = s_{e}+s_{o} = \frac{\log(2)}{4} +\frac{\pi}{8}\simeq 0.565986... \tag{4}$$

Answer 3

Hint: Notice that the sum is of consecutive odd powers of $x$ changing sign every two terms. What if you break it up into two sums, one of which is of the powers $4k+1$ and one of which is the powers of $4k+3$, both of alternating sign, and compare them to the Taylor series for $\frac{1}{1+y}$ for some suitable $y$.

Answer 4

Since you already received the answers, suppose that we compute the infinite sum first.

Notice that, using Euler's formula $$(-1)^\frac{k^2+k+2}{2}=-\cos \left( k (k+1)\frac{\pi}{2} \right)-i\sin \left( k (k+1)\frac{\pi}{2} \right)= -\cos \left( k (k+1)\frac{\pi}{2} \right)$$

The periodic sequence $$\{1,1,-1,-1,1,1,-1,-1,1,1,-1,-1,\cdots\}$$ corresponds to the expansion of $\frac{1+t}{1+t^2}$ but your exponent is $(2k-1)$ and start at $k=1$. Then, the infinite summation is just $$x\times\frac{1+x^2}{1+x^4}$$ Now $$I=\int_0^1 \frac{x(1+x^2)}{1+x^4}\,dx=\frac 12 \int_0^1 \frac{t+1}{t^2+1}\,dt=\frac 12\Bigg[\int_0^1 \frac{t}{ t^2+1}\,dt+\int_0^1 \frac{1}{ t^2+1}\,dt \Bigg]$$ $$I=\frac 12\Bigg[\frac{\log (2)}{2}+\frac \pi 4\Bigg]=\frac{\log(2)}{4} +\frac{\pi}{8}$$

Edit (for the asymptotics)

$$S_m= \sum_{k=1}^{m}(-1)^\frac{k^2+k+2}{2} x^{2k-1} =-\sum_{k=1}^{m}\cos \left( k (k+1)\frac{\pi}{2} \right)x^{2k-1}$$ $$S_m=-\frac{-x^6+x^4+x^2-1+x^8 \left(x^{8 \left\lfloor \frac{m-3}{4}\right\rfloor +6}-x^{8 \left\lfloor \frac{m-2}{4}\right\rfloor +4}-x^{8 \left\lfloor \frac{m-1}{4}\right\rfloor +2}+x^{8 \left\lfloor \frac{m}{4}\right\rfloor }\right)}{x \left(x^8-1\right)}$$ $$S_{4n-2}=\frac{x(1+x^2)}{1+x^4}+\frac{\left(1+x^2\right) x^{8 n-3}}{1+x^4}$$ $$J_n=\int_0^1 \frac{\left(1+x^2\right) x^{8 n-3}}{1+x^4}\,dx=\frac 12\int_0^1 \frac{(1+t) t^{4 n-2}}{1+t^2}\,dt$$ $$J_n=\frac{1}{8} \left(\psi \left(n+\frac{1}{4}\right)+\psi\left(n+\frac{1}{2}\right)-\psi (n)-\psi \left(n-\frac{1}{4}\right)\right)$$ For large values of $n$ $$J_n=\frac 1{8n}\Bigg[1+\frac{3}{8 n}+\frac{3}{32 n^2}+O\left(\frac{1}{n^4}\right) \Bigg]$$

For illustration $$J_{10}=\frac{6186754935233293}{10685862914126400}-\frac{\pi }{8}-\frac{\log (2)}{4}=0.01298046261$$ while the truncated series gives $$\frac{3323}{256000}= 0.01298046875$$