15
$\begingroup$

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,

$\endgroup$
0

2 Answers 2

21
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Excellent, this is very nice! Thanks. $\endgroup$ 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. $\endgroup$ May 12, 2011 at 23:54
14
$\begingroup$

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 $a_k=k!-\sum_{n=0}^k \frac{k!}{n!}\gamma_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.