To test convergence of improper integral $\int_{0}^{1} \left(\log\left(\frac{1}{x}\right)\right)^m\,\mathrm dx$

Question

To test convergence of improper integral

$$\int_{0}^{1} \left(\log\left(\frac{1}{x}\right)\right)^m\,\mathrm dx$$

I made cases and I am stuck on case in which I have to check convergence for $m< 0 , x =1$ is the point of infinite discontinuity of f on $[\frac{1}{2},1]$ after splitting integral at $1/2 .$ Problem is that I use series $\frac{1}{(1-x)^r}$ to apply comparison and I am getting

$$\lim_{x \to 1}(1-x)^{-r}(\log 1/x)^m$$

So, I am getting $r$ to be less than $0.$ But in textbook they say $m < 1\,.$

Answer 1

$$ \begin{align} \int_0^1\log\left(\frac1x\right)^m\,\mathrm{d}x &=\int_1^\infty\log(x)^m\,\frac{\mathrm{d}x}{x^2}\tag{1}\\ &=\int_0^\infty x^me^{-x}\,\mathrm{d}x\tag{2}\\[6pt] &=\Gamma(m+1)\tag{3} \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto\frac1x$
$(2)$: substitute $x\mapsto e^x$
$(3)$: Gamma integral

The integral in $(2)$ converges for all $m\gt-1$ and diverges for $m\le-1$.

Answer 2

Use the well-known inequality $(y-1)/y \leqslant \log y \leqslant y-1$ which is easily derived for $y \geqslant 1$ by bounding the integral $\log y = \int_1^y t^{-1} \, dt.$

For $0 < x \leqslant 1,$ we have

$$\log(1/x) \geqslant \frac{1/x-1}{1/x}= 1-x$$

If $m < 0$, then

$$[\log(1/x)]^{m} \leqslant (1-x)^m.$$

By the comparison test, the integral converges if $-1 < m < 0$.

Alternatively, using L'Hospital's rule

$$\lim_{x \to 1}\frac{\log (1/x)}{1-x} = 1,$$

and the limit comparison test applies. Hence, the integral converges if $-1 < m < 0$ and diverges if $m \leqslant -1$.