Any reference that we can find the following $$\Bigr[-\log(1-t)\Bigr]^x = t^x + x t^x \sum_{k=0}^\infty \psi_k(x+k)\,t^{k+1}; \quad \mbox{for all} \, x\in \mathbb R, \, |t|<1$$ where $\psi_k(.)$ are the Stirling Polynomials (Stirling Polynomials).

Thank in advance

  • $\begingroup$ this does not seem to be correct; just try $x=1$ and expand to second order in $t$; the left-hand-side becomes $-t-t^2/2$, while the right-hand-side becomes $t+t^2$, plus terms of order $t^3$; both the first and second order terms are different. $\endgroup$ Aug 26, 2016 at 17:32
  • $\begingroup$ It is a mistake. It is $\Bigr[-\log(1-t)\Bigr]^x $ not $\Bigr[\log(1-t)\Bigr]^x $. Thanks $\endgroup$
    – Z. Alfata
    Aug 26, 2016 at 17:52
  • $\begingroup$ the sign change fixes the first order term, but the second order term remains off by a factor $1/2$... $\endgroup$ Aug 26, 2016 at 18:36
  • 2
    $\begingroup$ @Carlo Beenakker: The formula below is hopefully correct... $\endgroup$
    – user111
    Aug 28, 2016 at 21:04
  • 3
    $\begingroup$ See Concrete Mathematics (2nd ed.) by Graham, Knuth, and Patashnik, formula (7.51), page 351, and formula (7.58), page 352. (They have a slightly different definition of Stirling polynomials.) $\endgroup$
    – Ira Gessel
    Sep 27, 2016 at 2:02

2 Answers 2


I edit my post to answer Carlo Beenakker's remark and also because I would like to add a reference, possibly more accurate than the two below. Theorem 7.1 p.13 of

A. Adelberg, A finite difference approach to degenerate Bernoulli and Stirling polynomials, Discrete Math. 140 (1995), 1-21,

states that, for $s$ a complex number, $$\left(\frac{(1+t)^{y}-1}{ty}\right)^{s}=\sum_{n=0}^{\infty}B_{n,s}(y)t^{n},$$ where $B_{n,s}(y)$ is a specific polynomial defined from divided differences of binomial coefficients. Letting $y$ tend to 0 gives $$(\log(1+t))^{s}=t^{s}\sum_{n=0}^{\infty}B_{n,s}(0)t^{n},$$ where it can be check that, with $\psi_{n}$ the Stirling polynomial, $$B_{n,s}(0)=\frac{(-1)^{n}}{n!}\frac{s}{s+n}\psi_{n}(s+n-1).$$ Hence, the seeked expansion at $t=0$ for a real power of $\log(1+t)$ is given by $$(\log(1+t))^{s}=t^{s}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\frac{s}{s+n}\psi_{n}(s+n-1)t^{n}.$$

The formula, written in a different form and with a complex exponent, appears on p.642 of this paper where generalized Stirling functions are used. These generalized Stirling functions are also studied in

Generalized Stirling Functions of Second Kind and Representations of Fractional Order Differences via Derivatives, P.L. BUTZER, A.A. KILBAS, J. TRUJILLO, Journal of Difference Equations and Applications, 2003 Vol. 9 (5), pp. 503–533

see in particular formula (3.9) there, to be compared with the definition of the Stirling polynomials. The link between the Stirling polynomials in your question and the generalized Stirling functions in these references shouldn't be too complicate to derive.


It should be remarked that the expansion is an immediate consequence of the Lagrange Inversion Formula, and of the definition of the Stirling polynomials via their generating function (as given e.g. in the Wolfram MathWord link you gave).

Let $f=\sum_{k=1}^\infty f_kt^k $ and $g=\sum_{k=0}^\infty g_kt^k $ be formal power series inverse of each other, with $f_1=g_1=1$; let $k\in\mathbb{N}$, $x\in\mathbb{C}$ and $y\in\mathbb{C}$ with $k+x+y=0$. Denoting as usual $[t^k]F$ the coefficient of $t^k$ in $F$, the Lagrange Inversion Formula may be written in the form: $$\frac{1}{x}[t^k]\left( \frac{f}{t}\right)^x+\frac{1}{y}[t^k]\left( \frac{g}{t}\right)^{y}=0$$ or in the equivalent, less symmetric form

$$[t^k]\left( \frac{f}{t}\right)^x=\frac{x}{x+k}[t^k]\left( \frac{t}{g}\right)^{k+x}.$$

If we take $f(t):= -\log(1-t) $ then its inverse is $g(t)=1-e^{-t}$, so

$$[t^k]\left( \frac{-\log(1-t)}{t}\right)^x=\frac{x}{x+k}[t^k]\left( \frac{t}{1-e^{-t}}\right)^{k+x}.$$ By the definition of the Stirling polynomials (with $x+k-1$ in place of $x$)

$$[t^k]\left( {t \over {1-e^{-t}}} \right) ^{x+k}= { S_k(x+k-1) \over k!} $$ so we get $$\big( -\log(1-t) \big)^x=xt^x \sum_{k=0}^\infty \frac{S_k(x+k-1)}{x+k}{t^k \over k!} $$

which is the expansion you want.

Incidentally, the expansion $$\log(-\log(1-t))=\log t-\sum_{k=1}^\infty {S_k(k-1)\over kk!}t^k,$$ can be obtained differentiating w.rto $x$ at $x=0$.


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