I need to see if the following integral converges:
$I= \int_1^{\infty} \frac{e^{-s}}{(s-1)^\alpha} ds \ , \ 0<\alpha \in \mathbb{R}$
What I did is divide it in two parts:
$ I = \int_1^{2} \frac{e^{-s}}{(s-1)^\alpha} ds + \int_2^{\infty} \frac{e^{-s}}{(s-1)^\alpha} ds$
the first one converges because the integrand is continuous and the interval is bounded. The second one converges because $ (s-1)^\alpha > 1$ for $s >2 $ so :
$ \int_2^{\infty} \frac{e^{-s}}{(s-1)^\alpha} ds < \int_2^{\infty} e^{-s}ds = e^{-2}$
From this I get that the integral must converge for all $\alpha$ considered.
The problem I have is that when I introduce it in WolframAlpha it adds a restraint in $\alpha$ i.e. $\alpha < 1$
Could you tell were I went wrong or why Wolfram gives that answer?
$$ \frac{e^{-s}}{(s-1)^{\alpha}} $$ is not bounded (and hence not continuous) on $[1,2]$: the denominator tends to zero. However, it is bounded by $e^{-1}(s-1)^{\alpha}$ there, since $e^{-s} \leqslant e^{-1}$ on $[1,2]$. But then $$ \int_{1+\varepsilon}^2 \frac{ds}{(s-1)^{\alpha}} = \frac{1}{1-\alpha}(1-\varepsilon^{1-\alpha}), $$ which only converges to a limit as $\varepsilon \downarrow 0$ if $\alpha<1$. The case $\alpha=1$ has to be checked separately (it still diverges, but logarithmically).