Analytically prove these Abel-Plana type integral equations 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,

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.

$$f(x)=(-1)^x \left(1-(x+1)^{\frac{1}{x+1}}\right)$$
$$h(x)=(-1)^x \left(x^{1/x}-1\right)$$
$$g(x)=x^{1/x}$$
$$\text{CMRB}=\sum _{n=0}^{\infty } f n=\sum _{n=1}^{\infty } f n=\sum _{n=1}^{\infty } h n$$
$$(MKB+\text{some n i}/\pi)=\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi  \text{it}} (g(t)) \, dt=\frac{i}{\pi }-i \int_0^{\infty } \frac{g (1+i t)}{e^{\pi  t}} \, dt, n\in{N}.$$
$$\underset{N\to \infty }{\text{lim}}\int_1^{2 N+1} e^{\pi  \text{it}} g(t) \, dt+i \int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi  t}-1} \, dt=\text{CMRB}$$
$$NoMKB+\Im{(MKB+\text{some n i}/\pi)}=i\int_0^{\infty } \frac{f (i t)-f (-i t)}{e^{2 \pi  t}-1} \, dt=\int_0^{\infty } \frac{i (h(1+i t)-h(1-i t))}{e^{2 \pi  t}-1} \, dt, n\in{N}.$$
You can see it worked out here.





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

