Series $\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$ Does anyone know the sums of the following two series?
$$\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$$
$$\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{4n-1}}{4n-1}$$
I encounter such series in my work.
 A: Well, to attack the first sum, if we call it $f(x)$, then
$$f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{4 n} = \frac1{1+x^4}$$
so that
$$f(x) = \int_0^x \frac{dt}{1+t^4}$$
You can use partial fractions to deduce that
$$\begin{align}f(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t+\sqrt{2}}{t^2+\sqrt{2} t+1}-\frac{t-\sqrt{2}}{t^2-\sqrt{2} t+1} \right ] \\ &= \frac1{2 \sqrt{2}} \int_0^x dt \left [\frac{t+1/\sqrt{2}+1/\sqrt{2}}{(t+1/\sqrt{2})^2+1/2}-\frac{t-1/\sqrt{2}-1/\sqrt{2}}{(t-1/\sqrt{2})^2+1/2} \right ] \\ &= \frac1{2 \sqrt{2}} \left [\arctan{(\sqrt{2} x+1)} - \frac{\pi}{4} + \arctan{(\sqrt{2} x-1)} + \frac{\pi}{4} \right ]\\ &+ \frac1{4 \sqrt{2}} \left [\log{(x^2+\sqrt{2} x+1)} - \log{(x^2-\sqrt{2} x+1)} \right ]\\ &=\frac1{2 \sqrt{2}} \arctan{\left (\frac{\sqrt{2} x}{1-x^2}\right )}+\frac1{4 \sqrt{2}} \log{\left (\frac{x^2+\sqrt{2} x+1}{x^2-\sqrt{2} x+1} \right )}\end{align} $$
The second sum is obviously similar, except that you are evaluating
$$g(x) = \int_0^x dt \frac{t^2}{1+t^4} $$
and, using similar steps, i.e., partial fractions, you get
$$\begin{align}g(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t}{t^2-\sqrt{2} t+1}-\frac{t}{t^2+\sqrt{2} t+1} \right ] \\ &= \frac1{2 \sqrt{2}} \int_0^x dt \left [\frac{t-1/\sqrt{2}+1/\sqrt{2}}{(t-1/\sqrt{2})^2+1/2}-\frac{t+1/\sqrt{2}-1/\sqrt{2}}{(t+1/\sqrt{2})^2+1/2} \right ]\\ &= \frac1{2 \sqrt{2}} \arctan{\left (\frac{\sqrt{2} x}{1-x^2}\right )}-\frac1{4 \sqrt{2}} \log{\left (\frac{x^2+\sqrt{2} x+1}{x^2-\sqrt{2} x+1} \right )} \end{align} $$
A: Consider the series
$$f(x) = \sum_{n=0}^{\infty} (-1)^n x^{4n} = \dfrac1{1+x^4}$$
then
$$\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{4n+1}}{4n+1} = \int_0^x f(t)dt = \int_0^x \dfrac{dt}{1+t^4} = \int_0^x \dfrac{dt}{(t^2+1)^2-(\sqrt2 t)^2}$$
and some brute force partial fraction to your rescue.
Proceed along similar lines for the second problem.
