Is $\int \limits _{0}^{1}x^\alpha|ln(x)|^\beta dx$ convergent?

Question

Is $\int \limits _{0}^{1}x^\alpha|ln(x)|^\beta dx$ convergent, where $\alpha >0, \beta >0$? Thanks .

Answer 1

Yes, since the Cauchy-Schwarz inequality grants: $$ \left(\int_{0}^{1}x^\alpha |\log x|^{\beta}\,dx\right)^2 \leq \frac{1}{2\alpha+1}\int_{0}^{+\infty} z^{2\beta}e^{-z}\,dz = \frac{\Gamma(2\beta+1)}{2\alpha+1}.$$ As a matter of fact, $$ \int_{0}^{1}x^\alpha|\log x|^\beta\,dx = \int_{0}^{+\infty}z^{\beta}e^{-(\alpha+1)z}\,dz = \frac{\Gamma(\beta+1)}{(\alpha+1)^{\beta+1}}$$ through the same substitution as above, $x=e^{-z}$.

Answer 2

Since $\log(x)\lt x$ for all $x\gt0$, it follows that, for $u\gt1$ and $\beta\gt0$, $$ \frac1{2^\beta\beta^\beta}\log(u)^\beta=\log\left(u^{\frac1{2\beta}}\right)^\beta\lt\left(u^{\frac1{2\beta}}\right)^\beta=u^{\frac12}\tag{1} $$ Using the substitution $x=\frac1u$ and $(1)$, we get $$ \begin{align} \int_0^1x^\alpha\left|\log(x)\right|^\beta\,\mathrm{d}x &\le\int_0^1\left|\log(x)\right|^\beta\,\mathrm{d}x\\ &=\int_1^\infty\log(u)^\beta\frac{\mathrm{d}u}{u^2}\\ &\le2^\beta\beta^\beta\int_1^\infty u^{\frac12}\frac{\mathrm{d}u}{u^2}\\ &=2^\beta\beta^\beta\int_1^\infty u^{-\frac32}\,\mathrm{d}u\\[3pt] &=2^{\beta+1}\beta^\beta\tag{2} \end{align} $$