I wish to find asymptotic equivalent function for $\sum_{k=100}^{n}\frac{k^{\alpha}}{\log\left(k\right)}$.
What I've tried so far is to use the fact that: $\int_{99}^{n}\frac{x^{\alpha}}{\log\left(x\right)}dx\leq\sum_{k=100}^{n}\frac{k^{\alpha}}{\log\left(k\right)}\leq\int_{100}^{n+1}\frac{x^{\alpha}}{\log\left(x\right)}dx$
But this integral is to hard to calculate.
Best Answer
Assume that $\alpha>-1$. Employing the Euler–Maclaurin formula, making a change of integration variables $t=\exp(s/(\alpha+1))$, and using the known asymptotic expansion of the exponential integral $\operatorname{Ei}$, we derive \begin{align*} \sum\limits_{k = 100}^n {\frac{{k^\alpha }}{{\log k}}} & = \int_{100}^n {\frac{{t^\alpha }}{{\log t}}\mathrm{d}t} + \mathcal{O}\!\left( {\frac{{n^\alpha }}{{\log n}}} \right) = \int_{(\alpha + 1)\log 100}^{(\alpha + 1)\log n} {\frac{{\mathrm{e}^s }}{s}\mathrm{d}s} + \mathcal{O}\!\left( {\frac{{n^\alpha }}{{\log n}}} \right) \\ & = \operatorname{Ei}((\alpha + 1)\log n) - \operatorname{Ei}((\alpha + 1)\log 100) + \mathcal{O}\!\left( {\frac{{n^\alpha }}{{\log n}}} \right) \\ & = \operatorname{Ei}((\alpha + 1)\log n) + \mathcal{O}\!\left( {\frac{{n^\alpha }}{{\log n}}} \right) \\ & \sim \frac{{n^{\alpha + 1} }}{{(\alpha + 1)\log n}}\sum\limits_{m = 0}^\infty {\frac{{m!}}{{(\alpha + 1)^m \log ^m n}}} +\mathcal{O}\!\left( {\frac{{n^\alpha }}{{\log n}}} \right) \\ & = \frac{{n^{\alpha + 1} }}{{(\alpha + 1)\log n}}\left[ {\sum\limits_{m = 0}^\infty {\frac{{m!}}{{(\alpha + 1)^m \log ^m n}}} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right] \\ &\sim \frac{{n^{\alpha + 1} }}{{(\alpha + 1)\log n}}\sum\limits_{m = 0}^\infty {\frac{{m!}}{{(\alpha + 1)^m \log ^m n}}} \end{align*} as $n\to +\infty$. Note that the lower limit of the sum does not alter the asymptotic expansion of the sum as long as it it fixed. Also, the asymptotic expansion agrees with that of $\operatorname{Li}(n^{\alpha + 1} )$.