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Greetings infinity series enjoyers. Not so long ago, I hypothesised existing of some equation connected with infinity summation, but could't prove it. The equation looks like $\sum_{k=x}^{\infty} f(k)=-2\pi \int_{-\infty}^{\infty}\frac {F (x-1/2+it)}{(e^{\pi t}+e^{-\pi t})^2}dt$ and primarly I was asking, if thats correct. Now a days I found a simple proof which is down below but when I know it's true, I have diferent questions. Does it has any value for mathematic? I mean equation is very general and works for divergent series too. Maybe it will find some use in mathematics. It is very similar to Abel-plana formula but that equationis look nicer for me and work for divergent series.

Proof

Let's firstly use Euler-Maclouren summation with some variation.

$\displaystyle \sum_{k=x}^{\infty} f (k)=- \left [ \frac {F (x)B_0}{0!}-\frac {f (x)B_1}{1!}+ \frac {f^{'}(x)B_2}{2!}- \frac {f^{''}(x)B_3}{3!}+ \frac {f^{'''}(x)B_4}{4!} ...\right]$

Now write Bernouli numbers in integral form in such way, that they will create Taylor series.

$\displaystyle \sum_{k=x}^{\infty} f (k)=-2 \pi \left [ \frac{F (x)}{0!}\int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it) ^0}{(e^{\pi t}+e^{-\pi t})^2}dt +\frac {f (x)}{1!} \int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it)^1}{(e^{\pi t}+e^{-\pi t})^2}dt + \frac {f^{'}(x)}{2!} \int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it)^2}{(e^{\pi t}+e^{-\pi t})^2}dt...\right]$

And just write Taylor series in form of function.

$\displaystyle \sum_{k=x}^{\infty} f (k)=-2 \pi \int_{-\infty}^{\infty} \frac { F (x) \frac{\left(-\frac {1}{2}+it\right)^0 }{0! }+ f (x) \frac {\left(-\frac {1}{2}+it\right)}{1!}+ f^{'}(x) \frac {\left(-\frac {1}{2}+it\right)^2}{2!}...}{(e^{\pi t}+e^{-\pi t})^2}dt =-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt $

Wreior
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  • What is the relationship between $f$ and $F$? Is $F$ some kind of transform of $f$ (e.g. Fourier, Laplace, or Mellin)? Or was $F$ a typo that should have been $f$? – Steven Clark Nov 08 '22 at 22:01
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    F it's just antideritative of f. But without +C term – Wreior Nov 09 '22 at 11:19
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    Without the assumption $x \in \mathbb Z$, which is not important, your formula becomes $$\frac 1 {2 \pi i} \int_{-1/2 - i \infty}^{-1/2 + i \infty} \pi^2 \csc^2( \pi s) , F(s + x) , ds = -\sum_{k = 0}^\infty \operatorname* {Res}_{s = k} , \pi^2 \csc^2( \pi s) , F(s + x).$$ This is false for $F(s) = 1$ (we cannot close the contour in such a way that the integral over the remainder of the contour is negligible; the sequence of integrals over $|s + 1/2| = n \land -\pi/2 < \arg(s + 1/2) < \pi/2$ tends to a finite non-zero limit). – Maxim Nov 09 '22 at 17:58
  • Is this a counterexample with $f(k)=e^{-k^2},x=1$? – Тyma Gaidash Jan 13 '23 at 19:20

1 Answers1

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Example ...
Take $f(u) = 1/u^2; x=1$. Then $$ \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} . \tag1$$ This conjecture tells us that $$ \frac{\pi}{2}\int_{-\infty}^\infty\frac{dt}{(\frac12+it)\cosh^2(\pi t)} = \frac{\pi^2}{6} . \tag2$$ Numerically, it works. Can we evaluate $(2)$? The imaginary part vanishes since it is odd, so $(2)$ reduces to $$ \frac{\pi}{4}\int_{-\infty}^\infty\frac{dt}{(\frac14+t^2)\cosh^2(\pi t)} = \frac{\pi^2}{6} . \tag3$$

Similarly $$ \pi\,\int_{-\infty }^{\infty }\!{\frac {1-4\,{t}^{2}}{ \left(1+ 4\,{t}^ {2} \right) ^{2} \cosh^2 \left( \pi\,t \right) }} \,{\rm d}t = \zeta(3) . $$

GEdgar
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