Obtaining a series expansion for an integral

Question

I am trying to obtain a series expansion for the integral $\int_0^{1/2} dx/(1+x^4)$ So far I used the fact for a Geometric series $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$ if $\lvert{r}\rvert<1$ and replaced $r$ with $-x^4$ and got $\frac{1}{1+x^4} = \sum_{k=1}^\infty (-1)^k x^{4k}$ if $\lvert{x}\rvert<1$

I am kind of unsure if this is even an answer or that if I need to go down a different path in order to solve this problem. If someone could tell me what to do in addition to this or what to differently or just show me how to do these types of problems it would be very appreciated. Thank you.

Answer 1

Yes, your approach us fine.

Integrating term-by-term, your integral equals $$\sum_{n\ge\color{red}{0}}\frac{(-1)^k}{4k+1}\frac1{2^{4k+1}}$$

I doubt if a nice closed form exists...

Answer 2

Too long for a side comment.

Considering $$\int_0^a \frac{dx}{1+x^4}=\sum_{k=0}^\infty (-1)^k\frac{a^{4k+1}}{4k+1}$$ it could be intersting to know in advance the number $n$ of terms to be summed in order to have $p$ significant correct figures.

This being said means that we want to know $n$ such that $$\frac{a^{4n+5}}{4n+5}\leq 10^{-p}$$ The solution for the equality is given in terms of Lambert function and write $$n=-\frac{5}{4}-\frac{W\left(-10^p \log (a)\right)}{4 \log (a)}$$ and we shall use $\lceil n\rceil$ for the calculations.

If, as in your case, $a<1$, we face a large argument, we can evaluate the value of Lambert function using the expansion given in the Wikipedia linked page $$W(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.

The table below shows the results for various $p$ $$\left( \begin{array}{ccc} p & \lceil n\rceil & \sum_{k=0}^n\frac{(-1)^k}{(4k+1)\,2^{4k+1}} \\ 1 & 0 & 0.50000000000000000000 \\ 2 & 0 & 0.50000000000000000000 \\ 3 & 1 & 0.49375000000000000000 \\ 4 & 2 & 0.49396701388888888889 \\ 5 & 2 & 0.49396701388888888889 \\ 6 & 3 & 0.49395762386485042735 \\ 7 & 4 & 0.49395807265276403029 \\ 8 & 5 & 0.49395804994623268729 \\ 9 & 6 & 0.49395805113832558279 \\ 10 & 6 & 0.49395805113832558279 \\ 11 & 7 & 0.49395805107409643972 \\ 12 & 8 & 0.49395805107762417674 \\ 13 & 9 & 0.49395805107742752924 \\ 14 & 10 & 0.49395805107743862064 \\ 15 & 10 & 0.49395805107743862064 \\ 16 & 11 & 0.49395805107743798904 \\ 17 & 12 & 0.49395805107743802530 \\ 18 & 13 & 0.49395805107743802320 \\ 19 & 14 & 0.49395805107743802332 \\ 20 & 14 & 0.49395805107743802332 \end{array} \right)$$

Edit

For $a=\frac 12$, quick and dirty linear regressions give $$n=0.779\,p-1.836 \qquad \text{and} \qquad p= 1.283\, n+2.362$$ showing, as robjohn commented, that "there should be approximately $1.2$ digits added for each term in the series".

Answer 3

Incomplete Beta Function $$ \begin{align} \int_0^{1/2}\frac{\mathrm{d}x}{1+x^4} &=\frac14\int_0^{1/16}\frac{x^{-3/4}\,\mathrm{d}x}{1+x}\\ &=\bbox[5px,border:2px solid #C0A000]{\operatorname{B}\left(\frac1{17};\frac14,\frac34\right)} \end{align} $$ Partial Fractions with $\alpha^2=i$; i.e. $\alpha=\frac{1+i}{\sqrt2}$ and $\alpha^3=\frac{-1+i}{\sqrt2}$ $$ \begin{align} \int_0^{1/2}\frac{\mathrm{d}x}{1+x^4} &=\int_0^{1/2}\frac12\left(\frac1{1-i^3x^2}+\frac1{1-ix^2}\right)\mathrm{d}x\\ &=\int_0^{1/2}\frac14\left(\frac1{1-\alpha^3x}+\frac1{1+\alpha^3x}+\frac1{1-\alpha x}+\frac1{1+\alpha x}\right)\mathrm{d}x\\ &=\frac1{4\alpha^3}\int_0^{\alpha^3/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\ &+\frac1{4\alpha}\int_0^{\alpha/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\ &=\frac1{4\alpha^3}\log\left(\frac{1+\alpha^3/2}{1-\alpha^3/2}\right)+\frac1{4\alpha}\log\left(\frac{1+\alpha/2}{1-\alpha/2}\right)\\ &=\frac{-1-i}{4\sqrt2}\log\left(\frac{2\sqrt2-1+i}{2\sqrt2+1-i}\right)+\frac{1-i}{4\sqrt2}\log\left(\frac{2\sqrt2+1+i}{2\sqrt2-1-i}\right)\\ &=\frac1{4\sqrt2}\log\left(\frac{5+2\sqrt2}{5-2\sqrt2}\right) +\frac1{2\sqrt2}\arctan\left(\frac{2\sqrt2}3\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\log\left(\frac{33+20\sqrt2}{17}\right)+\arctan\left(12\sqrt2\right)}{4\sqrt2}} \end{align} $$