What are some similarities between the MRB constant and its integrated analog?

Question

What are some similarities between MRB=$\sum _{x=0}^{\infty } e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) $ and M2= $\int_0^\infty e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) dx$ in their proper integrals from (0 to 1)?

Answer 1

I thought this similarity between their proper integrals was worth mentioning. You can see it worked and alter it here.

Answer 2

This is a perfect application for the Abel-Plana formula. The alternating series version would not work here, so the main definition of the formula will work.

$$\sum_{x\ge 1} f(x)=\frac{f(0)}2+\int_0^\infty f(x) dx+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}dx$$

Now let $f(x)=(-1)^x\left(1-\sqrt[x+1]{x+1}\right):$

$$\sum_{x\ge 1} (-1)^x\left(1-\sqrt[x+1]{x+1}\right) =\frac{(-1)^0\left(1-\sqrt[0+1]{0+1}\right)}2+\int_0^\infty (-1)^x\left(1-\sqrt[x+1]{x+1}\right) dx+i\int_0^\infty\frac{(-1)^{ix}\left(1-\sqrt[ix+1]{ix+1}\right)-(-1)^{-ix}\left(1-\sqrt[-ix+1]{-ix+1}\right)}{e^{2\pi x}-1}dx\implies \sum_{x\ge 1} (-1)^x\left(1-\sqrt[x+1]{x+1}\right) =\int_0^\infty (-1)^x\left(1-\sqrt[x+1]{x+1}\right) dx+i\int_0^\infty\frac{(-1)^{ix}\left(1-\sqrt[ix+1]{ix+1}\right)-(-1)^{-ix}\left(1-\sqrt[-ix+1]{-ix+1}\right)}{e^{2\pi x}-1}dx\implies i\int_0^\infty\frac{(-1)^{ix}\left(1-\sqrt[ix+1]{ix+1}\right)-(-1)^{-ix}\left(1-\sqrt[-ix+1]{-ix+1}\right)}{e^{2\pi x}-1}dx =\text M_2+\text{MRB}$$

Let me think of something that may not require difficult radicals. See the link for the theorem’s information. Please correct me and give me feedback!