Regarding this problem: For which values of $p,q$ does the integral $\int_0^1 x^p (\ln\frac{1}{x})^qdx$ converge?

Which comparison test can we use for the bound t→∞?


Let's change variable by $x = e^{-t},\ t>0$ $$\int_{0}^{1}x^p ln^q\frac{1}{x}dx = \int_{0}^{+\infty}e^{-t(p+1)}t^qdt $$

And divide last integral, for example, accordingly $t=1$ point. In 0-s neighbourhood we have same behaviour as for $t^q$, so integral converges when $q>-1$. In $+\infty$ behaviour is dictated by $e^{-t(p+1)}$ and, so, converges, when $p>-1$. It's easy to obtain by direct integrating: $$\begin{equation} \int\limits_1^{+\infty} e^{-t(p+1)} \mathrm{d} t \end{equation} = \dfrac{e^{-t(p+1)}}{-(p+1)} \Biggr|_{1}^{+\infty}$$

So, we need both: $q>-1$ and $p>-1$.

  • $\begingroup$ Thank you! but I'm not sure I understood how to direct integrating this incomplete integral $\endgroup$
    – Siv
    Jun 2 '20 at 21:43
  • $\begingroup$ Added line for direct integration. $\endgroup$
    – zkutch
    Jun 2 '20 at 23:39


If we put $$t=\frac 1x,$$

it will have the same nature as

$$\int_1^{+\infty}\frac{1}{t^p}(\ln(t))^q\frac{dt}{t^2}$$ or


this is a Bertrand integral (Google it).


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