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Our goal is evaluating: $$S=\sum_{n=0}^\infty(in)!= \sum\limits_{n=0}^\infty\Gamma(in+1) = 1.66159412484058… - 0.09593230517926 …i $$ which hypergeometric function generalizations cannot express. Attempting to rewrite it using $\Gamma(n)$’s integral representation does not work either as the geometric series converges for all $0\le x$. Even using a regularization and Cauchy principal value at $x=1$: $$S= \sum_{n=0}^\infty\int_0^\infty e^{-x}x^{i n}dx\mathop=^\text{reg}\int_0^\infty\frac{e^{-x}}{1-x^i}dx\mathop\approx^\text{PV}0.5-0.041380i$$ gives incorrect numerical results. The last idea was to convert it to a Barnes type integral, like with a Ramanujan interpolation based formula. With $\Gamma(in+1)$, the result diverges, but using $\Gamma(1-in)$:

$$\bar S=1+\frac i2\int_{c-i\infty}^{c+i\infty}((\cot(\pi z)+i)\Gamma(1-i z)dz\iff S=1-\frac i2\int_{ci-\infty}^{ci+\infty}(\coth(\pi z)-1)\Gamma(z+1)dz$$

Another useful formula may be $\Gamma(1-in)=-\pi i\frac{\operatorname{csch}(\pi n)}{\Gamma(i n)}$. Finally, Since $\lim_\limits{n\to\infty}\frac{2\ln(\text{Re}(\Gamma(in+1)))}{\pi n}=-1$, the sum’s real part roughly converges at $e^{-\frac\pi2n}$ speed. However, not much so far helps evaluate the sum.

What can we do?

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    $\begingroup$ I think at some point of your calculation you obtain an asymptotic series, which you evaluate directly or indirectly up to some big term, and the series should be evaluated instead up to the term with minimal absolute value, which is why your results are off numerically. Also, I confirm that $(in)!$ looks very confusing. $\endgroup$ Jan 26 at 15:48
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    $\begingroup$ I wonder if any alternate expressions for $\Gamma(\mathrm ix)$ exists for real $x$ $\endgroup$
    – K.defaoite
    Jan 26 at 21:01
  • $\begingroup$ Perhaps its complex characteristics would help $\endgroup$ Jan 26 at 21:14
  • $\begingroup$ Amusingly, the Calctastic calculator app for Android allows the computation of Gamma of complex values. Doing this up to (6i)! (which it evaluates as 0,0003587,,,-0.000341...) gives a sum up to there as 1.661566...-0.096028... $\endgroup$ Jan 27 at 4:33

1 Answer 1

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Consider the power series $$ F(z)=\sum_{n\geq 0}z^n\Gamma(i n+1). $$ Since $|\Gamma(i n+1)|\sim \sqrt{2\pi n}\,e^{-\frac \pi 2n}$ as $n\to+\infty$ (cf. DLMF), this defines a function analytic for $|z|<e^{\pi/2}$. In particular, $F(1)$ is the value we want to compute.

The following facts should be true.

  1. For $|z|<1$ we have the integral representation $$ F(z)=\int_0^{+\infty}\frac{e^{-s}}{1-zs^i}ds. $$

  2. We have $$ F(1)=\lim_{z\to 1_-}\int_0^{+\infty}\frac{e^{-s}}{1-zs^i}ds $$ where the notation indicates that the limit is taken as $z\to 1$ with $|z|<1$.

  3. This limit can be computed by the Sokhotski$-$Plemelji formula to be $$ F(1)=PV\int_0^{+\infty}\frac{e^{-s}}{1-s^i}ds+\pi \sum_{n\in\mathbb Z}\exp\bigl(-\exp(2\pi n)\bigr)\exp(2\pi n). $$

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  • $\begingroup$ Interestingly, taking the $PV$ at $x\in\Bbb N$ using the Abel Plana formula also gives the real part as $\frac\pi e+\frac12$, but it does not match $\text{Re}(S)=1.66 159\dots$ numerically $\endgroup$ Jan 26 at 19:47
  • $\begingroup$ let me update, I applied the formula but did not take into consideration all poles. I will update now (there should be numerical match now) $\endgroup$
    – Giulio R
    Jan 26 at 19:49
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    $\begingroup$ Sorry there was a misprint which I edited (the last factor is $exp(2 \pi n)$ rather than $exp(-2\pi n)$ as it was erroneously stated before. Now it should really match numerically. Let me know if the proofs of these three facts are necessary, if so I might add them when I have some time at some point in the future. $\endgroup$
    – Giulio R
    Jan 27 at 1:35
  • $\begingroup$ Corrected comment: The real part matches, but the imaginary part of $F(1)$ is $\frac12PV\int_0^\infty e^{-s}\cot(\frac{\ln(s)}2)ds$ with singularities at $s=e^{2\pi k},k\in\Bbb Z$ making it hard to calculate. $\endgroup$ Jan 27 at 14:01

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