I think I have started the following small set of proofs, but I could use a little help really proving them. $$ \text{Let } h(x)=(-1)^x \left(x^{1/x}-1\right).\text{CMRB}=\sum _{n=1}^{\infty } h(n).$$

$$\text{MKB}={\lim_{N \to \infty} }\int_1^{2 N} e^{i \pi t} t^{1/t} \, dt.$$

$$\text{Prove } \text{Re(MKB)}=\Im\left( \int_0^{\infty } \frac{h (1-i t)-h (1+i t)}{e^{2 \pi t}+1} \, dt\right)=\Im\left(\int_0^{\infty } \frac{h(1+i t)+h(1-i t)}{e^{2 \pi t}-1} \, dt\right).$$

$$\text{Let NoMKB}=\Im\left( \int_0^{\infty } \frac{h (1-i t)-h (1+i t)}{e^{2 \pi t}-1} \, dt\right).$$

$$\text{Prove NoMKB}+\text{Re(MKB)}=\text{CMRB}.$$ $$\text{Let} g(x)=x^{1/x}.$$ $$\text{Then prove Im(MKB)= }$$ $$-i{ \int _0^{\infty}\frac{g (1+i t)+g (1-i t)}{2 e^{\pi t}}}dt-\frac{i}{\pi }. $$

I know the proof that when $$g (x)=x^{1/x};\text{MKB}=-i \int_0^{\infty } \frac{g (1+i t)}{e^{\text{$\pi $t}}} \, dt-\frac{i}{\pi }.$$ It is here.

Also, how can we relate NoMKB to Im(MKB)to obtain MRB solely from MKB and vice-versa?

To see them worked out in Mathematica, go here. Similar integral proofs are cataloged here.

I got these from trial and error starting with the Abel-Plana formula, enter image description here from Wikipedia,

where, I think, $f(x)=(-1)^x(1-(1+x)^{1/(1+x)}),$ giving us

$$\text{ CMRB= }\sum _{n=0}^{\infty } f(n).$$ $$\text{MKB}={\lim_{N \to \infty} }\int_0^{2 N} f(x) dx.$$

Here is an outline of a "proof," but I will reward the bounty to a person who gives an analytic one. ![enter image description here $$f(x)=(-1)^x \left(1-(x+1)^{\frac{1}{x+1}}\right)$$ $$h(x)=(-1)^x \left(x^{1/x}-1\right)$$ $$g(x)=x^{1/x}$$ $$\text{CMRB}=\sum _{n=0}^{\infty } f n=\sum _{n=1}^{\infty } f n=\sum _{n=1}^{\infty } h n$$ $$(MKB+\text{some n i}/\pi)=\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi \text{it}} (g(t)) \, dt=\frac{i}{\pi }-i \int_0^{\infty } \frac{g (1+i t)}{e^{\pi t}} \, dt, n\in{N}.$$ $$\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi \text{it}} g(t) \, dt+i \int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi t}-1} \, dt=\text{CMRB}$$ $$NoMKB+\Im{(MKB+\text{some n i}/\pi)}=i\int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi t}-1} \, dt=\int_0^{\infty } \frac{i (h(1+i t)-h(1-i t))}{e^{2 \pi t}-1} \, dt, n\in{N}.$$

You can see it worked out here.

enter image description here enter image description here enter image description here enter image description here

  • $\begingroup$ What do you mean by analytic one? Without using Abel-Plana? $\endgroup$
    – Diger
    May 28, 2021 at 12:47
  • $\begingroup$ I'm afraid that my proof is lacking a lot of what I asked for. I hope that someone can prove it as if the Abel-Plana formula was not known. and thereby possibly prove some of the statements my "proof" left out. $\endgroup$ May 28, 2021 at 12:59
  • $\begingroup$ @Diger, I hope to post at mapleprimes.com/posts/… what I get. I plan on writing a paper summarizing those proofs. $\endgroup$ May 28, 2021 at 13:29

2 Answers 2


I'm not sure how you motivated yourself to this or how you progressed through it, but it feels very convoluted. Instead of working with your 3 functions $f,g,h$ let's just focus on $$f(x)=h(x+1)=e^{i\pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right)$$ (I picked the principal branch for $(-1)^x$) and in which case we obviously have by your definition $${\rm CMRB}=\sum_{n=0}^\infty f(n) \, .$$

Now you go on and define $${\rm MKB}=\int_1^{\infty} e^{i \pi t} t^{1/t} \, {\rm d}t$$ which is already a problem as it does not converge. You can make it converge by going over to $t^{1/t}-1$ instead of $t^{1/t}$. But then after substituting $t=x+1$ this is nothing else than $${\rm MKB} = \int_0^\infty f(x) \, {\rm d}x \, .$$

I think this form is much more suitable as you apply Abel-Plana. If you insist on using your limiting form for the integer $2N$ you can easily see they are connected

$$\lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} t^{1/t} \, {\rm d}t = \lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} (t^{1/t}-1) \, {\rm d}t + \lim_{N\rightarrow \infty} \int_1^{2N} e^{i\pi t} \, {\rm d}t = \int_0^\infty f(x) \, {\rm d}x - \frac{2i}{\pi} \, . $$

As for your "to prove" objective $${\rm NoMKB}+\Re ({\rm MKB})={\rm CMRB}$$ this follows immediately upon taking the real part of the Abel-Plana formula $$\sum_{n=0}^\infty f(n) = \frac{f(0)}{2} + \int_0^\infty f(x) \, {\rm d}x + i \int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1} \, {\rm d}t$$ since the LHS is real, $f(0)=0$ and for any complex number $z$ we have $\Re(iz)=-\Im(z)$.

Similarly you can take the imaginary part and use $\Im(iz)=\Re(z)$ so that $$\Im({\rm MKB})=\Re\left(\int_0^\infty \frac{f(-it) - f(it)}{e^{2\pi t}-1} \, {\rm d}t \right) \, .$$

Note that ${\rm NoMKB}$ is the imaginary part instead.

Regarding your other identity involving $\Re({\rm MKB})$ I think it should read $$\Re({\rm MKB}) = -\Re\left( \int_0^\infty \frac{f(it)+f(-it)}{e^{2\pi t}-1} \, {\rm d}t \right)$$ instead.


Thanks to Diger's answer, I believe this upmostly answers many parts of the question using the Abel-Plana formula. enter image description here


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