# 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.

• What do you mean by analytic one? Without using Abel-Plana? May 28, 2021 at 12:47
• 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. May 28, 2021 at 12:59
• @Diger, I hope to post at mapleprimes.com/posts/… what I get. I plan on writing a paper summarizing those proofs. May 28, 2021 at 13:29

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.

• Thank you. I'll study that, and I think I can get some good use out of it! May 30, 2021 at 1:23
• May 30, 2021 at 2:12

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