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

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)?

• Oct 11, 2021 at 13:44

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!

• Using the surds might make it harder than what the community and I have done using the Abel-Plana formula but it would be interesting to see what it produces. The results in my answer started from that formula. Oct 11, 2021 at 14:19
• Here is what we came up with: math.stackexchange.com/questions/4191171/… Oct 11, 2021 at 14:27
• There is also the definition of $$\text{MRB}=\sum_{x=0}^\infty(-1)^x\big(x^\frac1x-1\big)$$ as seen here. This one is different from the one in your question though. Oct 11, 2021 at 14:38
• I could have messed up; they're supposed to be the same since Exp(pi n i) =(-1)^n, and starting with n= 0 instead of 1 changes the sign of the summand. That is $\text{MRB}=\sum_{x=1}^\infty(-1)^x\big(x^\frac1x-1\big)$ Oct 11, 2021 at 14:55
• Tyma Gaidash , it could be that $\text{MRB}=\sum_1^\infty(-1)^x\big(\sqrt[x]x-1\big)$ not $\text{MRB}=\sum_0^\infty(-1)^x\big(\sqrt[x]x-1\big)$ Oct 11, 2021 at 20:30

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