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)?
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1$\begingroup$ This related question may help. $\endgroup$– Tyma GaidashOct 11, 2021 at 13:44
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2 Answers
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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!
answered Oct 11, 2021 at 14:02
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$\begingroup$ 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. $\endgroup$ Oct 11, 2021 at 14:19
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1$\begingroup$ Here is what we came up with: math.stackexchange.com/questions/4191171/… $\endgroup$ Oct 11, 2021 at 14:27
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$\begingroup$ 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. $\endgroup$ Oct 11, 2021 at 14:38
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$\begingroup$ 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) $ $\endgroup$ Oct 11, 2021 at 14:55
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1$\begingroup$ Tyma Gaidash , it could be that $\text{MRB}=\sum_1^\infty(-1)^x\big(\sqrt[x]x-1\big)$ not $\text{MRB}=\sum_0^\infty(-1)^x\big(\sqrt[x]x-1\big)$ $\endgroup$ Oct 11, 2021 at 20:30
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I thought this similarity between their proper integrals was worth mentioning. You can see it worked and alter it here.
