Convergence of $\int_0^{1/2}\frac{1}{x^\alpha \log x}dx$

Question

To establish, for which values of real parameter $\alpha$, the integral $$\int_0^{1/2}\frac{1}{x^\alpha \log x}dx$$ exists finite. For me, this problem is very difficult. Any suggestions please?

Answer 1

Your integral is convergent if and only if $\alpha<1$.

By the change of variable $$ x=e^{-u},\quad u=-\log x,\quad dx=-e^{-u}du, $$ one obtains $$ \int_{0}^{1/2} x^{-\alpha}\frac1{\log x} d \, x=-\int_{\ln 2}^{+\infty} \frac{e^{(\alpha-1)u}}{u}du $$ A potential problem of convergence is as $u \to +\infty$, in which case:

• if $\alpha -1<0$ then

$$ \frac{e^{(\alpha-1)u}}{u}<e^{(\alpha-1)u} $$ and the latter function is convergent over $[\ln 2,+\infty)$ giving the convergence of your initial integral.

• if $\alpha -1=0$ then

$$ \int_{0}^{1/2} x^{-\alpha}\frac1{\log x} d \, x=-\int_{\ln 2}^{+\infty} \frac{1}{u}du $$ giving the divergence of your initial integral.

• if $\alpha -1>0$ then

$$ \frac{e^{(\alpha-1)u}}{u} > \frac1u, \qquad u \to +\infty $$ giving the divergence of your initial integral.