Prove this series equal to this integral $\sum_{n=0}^\infty{\frac{(-1)^ny^{nx+1}}{nx+1}} = \int_{0}^y{\frac{dt}{1+t^x}}$

Question

I am answering question 1 here but am happy to get critiques on writing/mathematical correctness/ clarity. I have not yet answered question 2.

Question 1)

Prove the following:

$\forall x,y \in \mathbb{R}$ such that $|y^x|<1$

$\sum_{n=0}^\infty{\frac{(-1)^ny^{nx+1}}{nx+1}} = \int_{0}^y{\frac{dt}{1+t^x}}$

Question 2)

Can this be generalized to the complex numbers?

Answer 1

For a quick proof note that $$\frac{d}{dy}\sum_{n\geq 0} \frac{(-1)^ny^{nx+1}}{nx+1}=\sum_{n\geq 0} (-1)^ny^{nx}=\frac{1}{1+y^x}$$ when $|y^x|<1$, and $$\sum_{n\geq 0} \frac{(-1)^ny^{nx+1}}{nx+1}\biggr\rvert_{y=0}=\int_0^y\frac{1}{1+t^x}dt \biggr\rvert_{y=0},$$ so we have $$\sum_{n\geq 0} \frac{(-1)^ny^{nx+1}}{nx+1}=\int_0^y\frac{1}{1+t^x}dt$$ when $|y^x|<1$. This can be extended to $x,y\in\mathbb{C}$ since this function is analytic in our domain.

Answer 2

Question 1

I am answering my own question here but am happy to get critiques on writing/mathematical correctness/ clarity.

Here goes. We will establish the 3 equalities from left to right:

$\sum_{n=0}^\infty{\frac{(-1)^ny^{nx+1}}{nx+1}}%comments =\sum_{n=0}^\infty{\int_{0}^y{(-1)^n(t^x)^n}dt} =\int_{0}^y{\sum_{n=0}^\infty{(-1)^n(t^x)^n}dt} =\int_{0}^y{\frac{dt}{1+t^x}}$

The first equality. $\sum_{n=0}^\infty{\frac{(-1)^ny^{nx+1}}{nx+1}} =\sum_{n=0}^\infty{\int_{0}^y{(-1)^n(t^x)^n}}dt$

This is justified by standard calculus techniques: $\int_{0}^y(-1)^nt^{nx}dt= \frac{(-1)^nt^{nx+1}}{nx+1}|_0^y= \frac{(-1)^ny^{nx+1}}{nx+1}- \frac{(-1)^n0^{nx+1}}{nx+1}=\frac{(-1)^ny^{nx+1}}{nx+1}$

The second equality. $ \sum_{n=0}^\infty{\int_{0}^y{(-1)^n(t^x)^n}dt} =\int_{0}^y{\sum_{n=0}^\infty{(-1)^n(t^x)^n}dt} $

I can flip integral signs and summation signs. There should be no question at all for finite summands. This is sometimes referred to as the Tonelli theorems. Generally, $\sum\int f_n =\int\sum f_n$ for well-behaved $f_n$. I really should look into this deeper to get a stronger understanding. I think we need to know that $\int f_n$ and the $\sum f_n$ are both convergent and then we can get the equality.

The third equality. $\int_{0}^y{\sum_{n=0}^\infty{(-1)^n(t^x)^n}dt} =\int_{0}^y{\frac{dt}{1+t^x}}$

This holds because $\sum_{n=0}^{\infty}{(-t^x)^n}=\frac{1}{1+t^x}$. This is just a geometric series. Generally speaking $\sum_{n=0}^{\infty} r^n =\frac{1}{1-r} $ when $|r|<1$. So we should only use this tool when: $|t^x|<1$. So we can guarantee this by adding the requirement $|y^x|<1$. That's it! We have established the equality.

Question 2.

I am not sure. That's my next line of inquiry.