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I'm trying to follow this argument but with a slight difference: Continuation of series using Cahen-Mellin integral.

I start with $\Phi(s)=\sum_{n=1}^\infty e^{-n^s},$ which converges iff $s>0$ is real.

Applying the Mellin inversion theorem we have that:

$$e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$$

Where $K$ is a modified Bessel function of the second kind.

Then letting $x=e^{-n^{-s}}$ we have:

$$\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$

Am I on the right track? How do I get to the solution?

J. Zimmerman
  • 842
  • Can you please give a reference where you found the initial inverse Mellin transform $e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$? In the second inverse Mellin transform I don't believe $\sum_{n=1}^\infty e^{z,n^{-s}}$ converges. – Steven Clark Feb 07 '22 at 16:23
  • @StevenClark yes, I don't think it converges but then what did I do wrong? – J. Zimmerman Feb 07 '22 at 16:27
  • The Mellin Transform of e^(1/lnx) is $2K_1(2\sqrt(z))/\sqrt(z)$ – J. Zimmerman Feb 07 '22 at 16:29
  • or you just start by taking the inverse Mellin Transform of $\Gamma(z)\Gamma(z-1)$ which is the $K_1$ and then take one more inverse Mellin Transform of the $K_1$ – J. Zimmerman Feb 07 '22 at 16:33
  • The inverse Mellin transform $\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}x^{-z},dz=e^{\frac{1}{\ln(x)}}$ is only valid for $0<x<1$. In other words $\int_0^1 e^{\frac{1}{\ln(x)}},x^{z-1},dz=\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$ which is valid for $\Re(z)>0$, so $\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$ is the Mellin transform of $(\theta(x)-\theta(x-1)),e^{\frac{1}{\ln(x)}}$ versus the Mellin transform of $e^{\frac{1}{\ln(x)}}$. – Steven Clark Feb 07 '22 at 19:44
  • https://www.wolframalpha.com/input?i=inverse+Mellin+Transform+of+2BesselK%281%2C2sqrt%28z%29%29%2Fsqrt%28z%29 – J. Zimmerman Feb 07 '22 at 19:51
  • In Mathematica InverseMellinTransform[$\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$,z,x] generates the same result as WolframAlpha, but InverseMellinTransform[$\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$,z,x,GenerateConditions->True] clarifies the result is conditional for $0<x<1$. I also used Mathematica to obtain the integral result $\int_0^1 e^{\frac{1}{\ln(x)}},x^{z-1},dz=\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$. Mathematica can't seem to evaluate MellinTransform[$e^{\frac{1}{\ln(x)}}$,x,z]. – Steven Clark Feb 07 '22 at 20:06
  • In fact Mathematica indicates the Mellin Transform integral $\int_0^\infty e^{\frac{1}{\ln(x)}},x^{z-1},dx=\frac{2,K_1(2\sqrt{z})}{\sqrt{z}}$ doesn't converge. – Steven Clark Feb 07 '22 at 20:14
  • @StevenClark okay, so $0<x<1.$ and $\Re(z)>0.$ – J. Zimmerman Feb 07 '22 at 20:14
  • I think you mean $dx$ at the end of the integral – J. Zimmerman Feb 07 '22 at 20:15
  • what impact does this have on the evaluation of the last integral $\Phi(s)$? – J. Zimmerman Feb 07 '22 at 20:15
  • Yes I corrected it. I'm wondering if perhaps $\sum_{n=1}^\infty e^{z,n^{-s}}$ converges in a distributional sense. For example the Fourier series for the Dirac comb converges in a distributional sense (see https://en.wikipedia.org/wiki/Dirac_comb#Fourier_series). – Steven Clark Feb 07 '22 at 20:16
  • If $\sum_{n=1}^\infty e^{z,n^{-s}}$ converges in a distributional sense, then one could perhaps use the distributional representation to evaluate the related inverse Mellin transform integral. – Steven Clark Feb 07 '22 at 20:21
  • okay. the substitution $x=e^{-n^{-s}}$ is valid right? – J. Zimmerman Feb 07 '22 at 20:26
  • Since your initial inverse Mellin transform integral is only valid for $0<x<1$, it seems to me any result you're able to obtain will also only be valid for $0<x<1$ instead of $x>0$ (which is better than nothing), but it's not obvious to me how to proceed with evaluation of the second inverse Mellin transform integral at this point. – Steven Clark Feb 07 '22 at 21:15
  • from the link(in the post), we know the answer right? Is it arriving at that answer that's difficult? – J. Zimmerman Feb 07 '22 at 21:36
  • Yes that's one formula, but I wonder if this question leads to another formula. I mentioned above the initial inverse Mellin transform is only valid for $0<x<1$. After thinking about it some more it seems to me the substitution $x=e^{-n^{-s}}$ leads to $0<e^{-n^{-s}}<1$ which is valid for all $s>0$ (since $n\in\mathbb{Z}\land n>0$), so perhaps the result of the second inverse Mellin transform would be valid for all $s>0$. It appears the formula from the answer in your initial question is valid for $0<\Re(s)<1$, so perhaps the approach here would result in a different formula. – Steven Clark Feb 08 '22 at 16:35
  • can Mathematica evaluate the inverse Mellin transform of $e^{\frac{1}{\log(z)}}$? – J. Zimmerman Feb 10 '22 at 20:14
  • No Mathematica gives up pretty much immediately as does Wolfram Alpha, but I'm not sure how $\mathcal{M}_x^{-1}\lefte^{\frac{1}{\log (x)}}\right$ is related to this question. – Steven Clark Feb 10 '22 at 20:46
  • @StevenClark If the substition is valid, then why would the integral not converge? What I'm saying is why did this approach work in the linked post but in this case there's no convergence of the integral? – J. Zimmerman Oct 29 '22 at 20:16
  • Don't know how to help to your question, but I would like to note something: if $z$ and $n$ are complex variable then $ \left(e^z\right)^n \neq e^{nz}$ since there is a multiplicity of values, be cautious about it - see Wiki – Joako Feb 07 '23 at 17:16
  • $n$ is a natural number – J. Zimmerman Feb 11 '23 at 00:26

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