For what real values of $a$ the following integral converges?
$$ \int_0^1 (-\ln{x})^a dx $$
Do some u-subbing. $$\int_0^1 (-\ln(x))^a\, dx = \int_{-\infty}^0 (-\ln(e^x))^a e^x\,dx = \int_0^\infty x^a e^{-x}\, dx. $$ This integral converges provided $a > -1$.
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Do some u-subbing. $$\int_0^1 (-\ln(x))^a\, dx = \int_{-\infty}^0 (-\ln(e^x))^a e^x\,dx = \int_0^\infty x^a e^{-x}\, dx. $$ This integral converges provided $a > -1$.