Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:36:43Z http://mathoverflow.net/feeds/question/64556 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64556/evaluating-the-integral-int-1-infty-frac-u-u2-left-log-u-right Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$ Eric Naslund 2011-05-11T01:26:13Z 2012-04-16T11:41:08Z <p>I am trying to find a formula for the following integral for non-negative integer $k$: </p> <p><code>$$\int_1^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$$</code></p> <p>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 <a href="http://en.wikipedia.org/wiki/Stieltjes_constants" rel="nofollow">Stieltjes Constants</a>. 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.</p> <p>Thanks a lot,</p> http://mathoverflow.net/questions/64556/evaluating-the-integral-int-1-infty-frac-u-u2-left-log-u-right/64562#64562 Answer by Julian Rosen for Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$ Julian Rosen 2011-05-11T03:29:14Z 2011-05-11T03:29:14Z <p>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 $a_k=k!-\sum_{n=0}^k \frac{k!}{n!}\gamma_n$.</p> http://mathoverflow.net/questions/64556/evaluating-the-integral-int-1-infty-frac-u-u2-left-log-u-right/64568#64568 Answer by Qiaochu Yuan for Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$ Qiaochu Yuan 2011-05-11T05:46:16Z 2011-05-11T05:51:22Z <p>Let $a_k$ be the integral. Then</p> <p>$$\begin{eqnarray*} \sum_{k \ge 0} \frac{a_k}{k!} t^k &amp;=&amp; \int_1^{\infty} \frac{ \{ u \} }{u^2} e^{t \log u} \, du \\ &amp;=&amp; \int_1^{\infty} \{ u \} u^{t-2} \, du \\ &amp;=&amp; \frac{1 - \zeta(1 - t) - \frac{1}{t}}{1 - t} \\ &amp;=&amp; \frac{1}{1 - t} \left( 1 - \sum_{n \ge 0} \frac{\gamma_n}{n!} t^n \right). \end{eqnarray*}$$</p> <p>(Generating functions are good for more than combinatorics!) This is equivalent to Julian Rosen's answer, but (I think) packaged slightly more conveniently. </p>