Divergence of $\int_{0}^{1/2}\frac{x^s-1}{-\log(x)}$

Question

I am interested in an integral $\int_{0}^{1/2}\frac{x^s-1}{-\log(x)}$ for $s \le -1$.

The lecture notes I am using claim that it diverges. I am not sure why.

I was trying to find a function $0 \le g \le \frac{x^s-1}{-\log(x)}$ on $(0,\frac{1}{2}]$ such that $\int_{0}^{1/2}g= \infty$. The only thing which comes to mind is $g=\frac{x^s-1}{\frac{1}{x}-1}$, but this apparently does not work e.g. for $s = -1$ (though seems to work for $s \le -2$).

Answer 1

Hint. Note that $$\int_{0}^{1/2}\frac{x^s-1}{-\log(x)}\,dx =\int_{0}^{1/2}\frac{1}{x^{-s}(-\log(x))}\,dx+\int_{0}^{1/2}\frac{1}{\log(x)}\,dx\\\geq \int_{0}^{1/2}\frac{1}{x(-\log(x))}\,dx-\frac{1}{2\log(2)}.$$ Moreover $$\int \frac{1}{x\log(x)}\,dx=\log(|\log(x)|)+C.$$ Can you take it from here?

Answer 2

When $s\le-1$, we have $$\frac{x^{s}-1}{-\log x}\ge\frac{x^{-1}-1}{-\log x}\ge\frac{1}{-2x\log x},$$ since $0<x<\frac{1}{2}$.

The antiderivative of the right-handed side is known, i.e. $$\int\frac{1}{-2x\log x}\mathrm{d}x= -\frac12\log|\log x|+C.$$ Hence, the integral diverges.

Answer 3

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With $\ds{\pars{% \begin{array}{rcl} \ds{x} & \ds{=} & \ds{\expo{-t}} \\ \ds{t} & \ds{=} & \ds{-\ln\pars{x}} \end{array}}}$ and $\ds{\pars{s + 1 > 0 \implies s > - 1}}$:

\begin{align} \int_{0}^{1/2}{x^{s} - 1 \over -\ln\pars{x}}\,\dd x & = \int_{\infty}^{-\ln\pars{1/2}}{\expo{-st} - 1 \over t}\,\pars{-\expo{-t}}\dd t = \int_{\ln\pars{2}}^{\infty}{\expo{-\pars{s + 1}t} - \expo{-t} \over t}\,\dd t \\[1cm] & = -\ln\pars{\ln\pars{2}}\bracks{\expo{-\pars{s + 1}\ln\pars{2}} - \expo{-\ln\pars{2}}} \\[2mm] & -\int_{\ln\pars{2}}^{\infty}\ln\pars{t}\braces{\vphantom{\Large A}% \expo{-\pars{s + 1}t}\left[\vphantom{\large A}-\pars{s + 1}\right] - \expo{-t}\pars{-1}}\dd t \\[1cm] & = -\ln\pars{\ln\pars{2}}\pars{{1 \over 2^{s + 1}} - {1 \over 2}} + \int_{\pars{s + 1}\ln\pars{2}}^{\infty}\ln\pars{t \over s + 1}\expo{-t}\,\dd t \\[2mm] & -\int_{\ln\pars{2}}^{\infty}\ln\pars{t}\expo{-t}\,\dd t \\[5mm] & = -\ln\pars{\ln\pars{2}}\pars{{1 \over 2^{s + 1}} - {1 \over 2}} - {\ln\pars{s + 1} \over 2^{s + 1}} + \int_{\pars{s + 1}\ln\pars{2}}^{\ln\pars{2}}\ln\pars{t}\expo{-t}\,\dd t \\[5mm] & \stackrel{\mrm{as}\ s\ \to\ \pars{-1}^{+}}{\sim}\,\,\, \bbx{-\ln\pars{s + 1} - {\ln\pars{\ln\pars{2}} \over 2} + \int_{0}^{\ln\pars{2}} \ln\pars{t}\expo{-t}\,\dd t} \end{align}