I am trying to find a formula for the following integral for non-negative integer $k$:

$$\int_1^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$$

My first thought was to use the formula $$\zeta(s)-\frac{1}{s-1}=1-s\int_1^\infty u^{-s-1}\{u\}du$$ where $\{u\}$ refers to the fractional part. We can then take derivatives with respect to $s$ and use the Laurent expansion for $\zeta(s)$. It follows that each integral must be a finite linear combination of the Stieltjes Constants. All of the coefficients must be integers, and $\gamma_n$ can only appear if $n\leq k$. (This checks out numerically for $k=0,1,2$) Unfortunately, I am not sure what the pattern is, but I feel these particular integrals must be very common, and must have been dealt with before. I am hoping someone can give me a reference, or give a solution.

Thanks a lot,


2 Answers 2


Let $a_k$ be the integral. Then

$$\begin{eqnarray*} \sum_{k \ge 0} \frac{a_k}{k!} t^k &=& \int_1^{\infty} \frac{ \{ u \} }{u^2} e^{t \log u} \, du \\\ &=& \int_1^{\infty} \{ u \} u^{t-2} \, du \\\ &=& \frac{1 - \zeta(1 - t) - \frac{1}{t}}{1 - t} \\\ &=& \frac{1}{1 - t} \left( 1 - \sum_{n \ge 0} \frac{\gamma_n}{n!} t^n \right). \end{eqnarray*}$$

(Generating functions are good for more than combinatorics!) This is equivalent to Julian Rosen's answer, but (I think) packaged slightly more conveniently.

  • $\begingroup$ Excellent, this is very nice! Thanks. $\endgroup$ Commented May 11, 2011 at 15:27
  • $\begingroup$ This technique is very useful. I have already used it for 2 other problems. I'll keep generating series in mind in the future. $\endgroup$ Commented May 12, 2011 at 23:54

Write $a_k$ for your integral. If we define $g(s)=\zeta(s)-\frac{1}{s-1}$, then $\left(\frac{d}{ds}\right)^n|_{s=1}g(s)=(-1)^n\gamma_n$. Your observation can be written $a_k=(-1)^k\left(\frac{d}{ds}\right)^k|_{s=1}\left(\frac{1}{s}-\frac{1}{s}g(s)\right)$. The derivative can be computed directly to give $a_k=k!-\sum_{n=0}^k \frac{k!}{n!}\gamma_n$.


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