More about integration by parts and the Möbius function: the product $\prod_{n=1}^\infty\left(1+x^n\right)^{\mu(n)/n}$

Question

For $|x|<1$ let (I believe that it converges) $$f(x)=\prod_{n=1}^\infty\left(1+x^n\right)^{\frac{\mu(n)}{n}},$$ where $\mu(n)$ is the Möbius function. Using the infinite product that satisfies the Möbius function, see $(16)$ from this MathWorld, I am saying the specialization at $x^2$ and the integration by parts $$u=f(x),\qquad dv=e^{-x}dx$$ thus for $$\int_0^1\left(\prod_{n=1}^\infty\left(1+x^n\right)^{\frac{\mu(n)}{n}}\right)e^{-x}dx$$ I've written this

Claim. The identity $$\frac{\sqrt{2}}{\pi}\operatorname{erf}(1)-\left(1-\frac{1}{e}\right)=\int_0^1 e^{-x}f(x)\left(\sum_{n=1}^\infty\mu(n)\frac{x^{n-1}}{1+x^n}\right)dx$$ holds.

Question 1. I've the calculations that I did, thus only is required if you can justify the convergence issues and tell me if such Claim is right or wrong.

Motivation of previous question, and related remarks. I try to explore what calculations are feasibles with this infinite product. The product $1=\prod_{n=1}^\infty 2^{\mu(n)/n}$ follows from the Prime Number Theorem. I had failures in a subsequent integration by parts from the deduced one in Claim.

Question 2. Can you provide me hints, details to get for integers $n\geq 1$ $$\int \frac{e^{-x}x^{n-1}}{1+x^n}dx,$$ and $$\int_0^1 \frac{e^{-x}x^{n-1}}{1+x^n}dx$$ from particular values of special functions? Thanks a lot.

I know from an online calculator that we can calculate simple examples of the definite integral. If you know where refers the literature this integral, also feel free to add it, as you see it.

Answer 1

Applying formal manipulations, by the Taylor series of $\log(1+x)$ we have:

$$f(x)=\exp\sum_{n\geq 1}\frac{\mu(n)}{n}\sum_{m\geq 1}\frac{(-1)^{m+1} x^{mn}}{m}=\exp\left(-\sum_{h\geq 1}\frac{x^h}{h}\left(\mu*(-1)^{\cdot}\right)(h)\right) $$ If $h$ is an odd integer greater than one, $\left(\mu*(-1)^{\cdot}\right)(h) = -\left(\mu*1\right)(h)=0$.
If $h=2^k u$ with $u$ odd, $$\left(\mu*(-1)^{\cdot}\right)(h) = \sum_{d\mid u}\mu(d)-\sum_{d\mid u}\mu(2d)$$ equals zero unless $u=1$ and $k=1$. It follows that $\prod_{n\geq 1}\left(1+x^n\right)^{\frac{\mu(n)}{n}}$ is just a convoluted way for writing $\color{red}{\exp\left(x-x^2\right)}$. Similarly, $\sum_{n\geq 1}\mu(n)\frac{x^{n-1}}{1+x^n}$ is just a convoluted way for writing the logarithmic derivative of the previous function, namely $\color{red}{1-2x}$. After the syntactic sugar given by $\mu(n)$ is entirely removed, your questions boil down to simple calculus questions.