Function that has per MacLaurin series $\sum_{n=0}^\infty \frac{x^{4n+1}}{4n+1}$

Question

What is the function of the MacLaurin series:

$$\sum_{n=0}^\infty\frac{x^{4n+1}}{4n+1}$$

I think is a complex function, but I have no idea how to start. Is it already discovered?

Answer 1

See that $$f'(x)=\sum_{n=0}^\infty x^{4n}$$ This is just a geometric series, with common ratio $x^4$ and hence provided $|x^4|<1$ we get $$f'(x)=\frac{1}{1-x^4}$$ Hence $$f(x)=\int_0^x f'(s)\mathrm ds=\int_0^x\frac{1}{1-s^4}\mathrm ds$$ As given by Guillermo.

Introduce a partial fraction decomposition: $$\frac{1}{1-s^4}=\frac{1}{(1+s^2)(1-s^2)} \\ =\frac{A}{1+s^2}+\frac{B}{1-s^2}$$ We see that $$(1+s^2)B+(1-s^2)A=1 \\ B+A+s^2(B-A)=1$$ This means that we must have $B=A=1/2$. Therefore

$$f(x)=\frac{1}{2}\int_0^x\frac{1}{1+s^2}\mathrm ds+\frac{1}{2}\int_0^x \frac{1}{1-s^2}\mathrm ds$$ Which, assuming $x\in(0,1)$, is $$\boxed{f(x)=\frac{1}{2}\big(\arctan(x)+\operatorname{arctanh}(x)\big)}$$

Answer 2

This function is $$f(x) =\int_0^x\frac{1}{1-t^4}dt.$$ Why? Give it a try.