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http://mathhelpboards.com/math-note...polylogarithms-associated-functions-6520.html

- Thread starter MarkFL
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http://mathhelpboards.com/math-note...polylogarithms-associated-functions-6520.html

- Feb 13, 2012

- 1,704

Honestly all that seems to me a little questionable... the series $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k}$ diverges so that writing $\displaystyle a = \sum_{k=1}^{\infty} \frac{1}{k}$ or something like that is a nonsense. The connection between tha digamma function and the Riemann zeta function do exist and I 'discovered' it some years ago...Incidentally, it is occasionally convenient to express this series in terms of the Zeta function \(\displaystyle \zeta(s)\,\)

\(\displaystyle \zeta(s)=\sum_{k=1}^{\infty}\frac{1}{k^s}\)

In particular, since

\(\displaystyle \zeta(1)=\sum_{k=1}^{\infty}\frac{1}{k}\)

We can re-write the polygamma function series as

\(\displaystyle \psi_0(x)=\zeta(1)-\gamma-\sum_{k=1}^{\infty}\frac{1}{k-1+x}=\zeta(1)-\gamma-\sum_{k=0}^{\infty}\frac{1}{k+x}

\)

Note that the zeta function is divergent at 1 - where it has a simple pole, and similarly, the infinite series in x that's subtracted from it also diverges, but at a different rate. The end result is that, for finite x not equal to zero or a negative integer, the difference of these two series is a finite limit.

There are two different definitions of the digamma function, and precisely...

$\displaystyle \psi(x) = \frac{d}{dx} \ln \Gamma(x),\ \Gamma(x) = \int_{0}^{\infty} t^{x-1}\ e^{- t}\ dt\ (1)$

... and...

$\displaystyle \phi(x) = \frac{d}{dx} \ln x!,\ x! = \int_{0}^{\infty} t^{x}\ e^{- t}\ dt\ (2)$

Although (1) is more common I prefer (2) but in any case the diffence is minimal because is $\displaystyle \phi(x)=\psi(x+1)$. Starting from the Weierstrass product...

$\displaystyle \frac{1}{x!}= e^{\gamma x}\ \prod _{n=1}^{\infty} (1 + \frac{x}{n}) e^{- \frac{x}{n}}\ (3)$

... with symple steps we obtain...

$\displaystyle \lambda(x) = \ln x! = - \gamma\ x - \sum_{n=1}^{\infty} \{\ln (1+\frac{x}{n}) - \frac{x}{n}\}\ (4)$

Now we are interested to the Taylor expansion of $\displaystyle \lambda(x)$ around x=0 that is...

$\displaystyle \lambda(x) = \sum_{k=0}^{\infty} \frac{\lambda^{(k)} (0)}{k!} x^{k}\ (5)$

The coeffcients are computed deriving (4)...

$\displaystyle \lambda^{\ '} (x) = - \gamma - \sum_{n=1}^{\infty}\{\frac{1}{n} (1+ \frac{x}{n})^{-1} - \frac{1}{n}\} \implies \lambda^{\ '} (0) = - \gamma$

$\displaystyle \lambda^{\ ''} (x) = \sum_{n=1}^{\infty} \frac{1}{n^{2}} (1 + \frac{x}{n})^{-2} \implies \lambda^{\ ''} (0) = \zeta(2)$

... and proceeding we obtain the general result...

$\displaystyle \lambda^{\ (k)} (0) = (-1)^{k} (k-1)!\ \zeta(k)\ (6)$

... so that we arrive to write...

$\displaystyle \lambda(x) = - \gamma\ x + \sum_{k=2}^{\infty} (-1)^{k} \zeta(k) \frac{x^{k}}{k}\ (7)$

... and finally...

$\displaystyle \phi(x) = \frac{d}{d x} \lambda (x) = - \gamma + \sum_{k=2}^{\infty} (-1)^{k} \zeta(k)\ x^{k-1}\ (8)$

Kind regards

$\chi$ $\sigma$

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- Sep 16, 2013

- 337

\(\displaystyle \int_0^{\infty}\frac{p(x)-q(x)}{x}\,dx=???\)

say, where if separated, the integrals

\(\displaystyle \int_0^{\infty}\frac{p(x)}{x}\,dx\)

and

\(\displaystyle \int_0^{\infty}\frac{p(x)}{x}\,dx\)

are both divergent, even though the difference might well be a finite limit, similar to the series I used.

Also, I just meant that occasionally, when dealing with differences of certain polygamma-related series, it's notationally convenient to write them in this simpler form...

- Jan 17, 2013

- 1,667

Great tutorial , marked on my reading list !