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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?

Mason
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  • Seems very similar to the form in this question. I wonder if we can get a closed form for when the denominator is raised to a power. – Mason Dec 21 '18 at 03:23

2 Answers2

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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.

Jacob
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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.

Mason
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    Please include this as part of your question, and clarify that you're looking for [tag:proof-verification]. Also, please note that it isn't a complete answer due to the last part. –  Apr 14 '18 at 22:22
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    Also, for an actual critique on your writing: Don't include nonsense about "dem da rules" and "god given right[s]." –  Apr 14 '18 at 22:23
  • Good mathematical writing is easy to understand. "Dem da rules" is vague, imprecise, and annoying to come across when reading an otherwise serious thing. Good writing doesn't need to be dry, but it also shouldn't have nonsense for the sake of nonsense. –  Apr 14 '18 at 22:29