Problem - G. Apostolopoulos (Greece)
Prove
\[\sum_{k=2}^\infty\frac{1}{k}\int_0^1\left\{\frac{1}{\sqrt[k]{x}}\right\}\,\mathrm dx=\gamma\]where $\gamma$ is the Euler-Mascheroni constant.
Solution
First we let $x=u^{-k}$ so that $\mathrm dx=-ku^{-k-1}\,\mathrm du$ and
\[\sum_{k=2}^\infty\frac{1}{k}\int_0^1\left\{\frac{1}{\sqrt[k]{x}}\right\}\,\mathrm dx=\sum_{k=2}^\infty\frac{1}{k}\int_\infty^1\left\{u\right\}\left(-ku^{-k-1}\right)\,\mathrm du.\]We can then turn the integral into something more manageable,
\[\begin{align*} \sum_{k=2}^\infty\int_1^\infty\left\{u\right\}u^{-k-1}\,\mathrm du &= \int_1^\infty\frac{\left\{u\right\}}{u}\sum_{k=2}^\infty\frac{1}{u^k}\,\mathrm du, \\ &\qquad\text{by monotone convergence} \\ &= \int_1^\infty\frac{\left\{u\right\}}{(u-1)u^2}\,\mathrm du.\end{align*}\]Then,
\[\begin{align*} \int_1^\infty\frac{\left\{u\right\}}{(u-1)u^2}\,\mathrm du &= \sum_{n=1}^\infty\int_n^{n+1}\frac{\left\{u\right\}}{(u-1)u^2}\,\mathrm du \\ &= \sum_{n=1}^\infty\int_0^1\frac{u}{(u+n-1)(u+n)^2}\,\mathrm du,\;\;\text{by letting }u\mapsto u+n \\ &= \int_0^1 u\sum_{n=1}^\infty\frac{1}{(u+n-1)(u+n)^2}\,\mathrm du, \\ &\qquad\text{by uniform convergence} \\ &= \int_0^1 u\sum_{n=1}^\infty\left(\frac{1}{u+n-1}-\frac{1}{u+n}-\frac{1}{(u+n)^2}\right)\,\mathrm du \\ &= \int_0^1 u\left(\frac{1}{u}+\frac{1}{u^2}-\psi'(u)\right)\,\mathrm du, \\ &\qquad\text{where }\psi'(u)\text{ is the trigamma function} \\ &= 1+\int_0^1\left(\frac{1}{u}-u\psi'(u)\right)\,\mathrm du \\ &= 1+\left[\ln(u)+\ln\left(\Gamma(u)\right)-u\psi(u)\right]\vert_0^1 \\ &= 1+\gamma-1 \\ &= \gamma\end{align*}\]as was to be shown.