I have a question here, which I would appreciate some help for: for which values of $\alpha$ is the improper integral $\int_0^1\frac{e^x - 1}{x^\alpha}dx$ convergent?
I kind of get that I'm supposed to convert it into a power series using power series of $e^x$ but I'm stuck with how to continue from there. Can someone please explain the full steps and logic behind it?
On
Hint: near zero the integrand behaves as
$$ \frac{e^x-1}{x^\alpha}\sim \frac{x}{x^\alpha}. $$
See related techniques.
Hint
As you properly suspected, you need to expand $e^x$ first as a power series. Since $$e^x=\sum_{i=0}^{\infty} \frac{x^i}{i!}$$ then $$\frac{e^x-1}{x^{\alpha}}=\frac{1}{x^{\alpha}}\sum_{i=1}^{\infty} \frac{x^i}{i!}=\sum_{i=1}^{\infty} \frac{x^{i-\alpha}}{i!}=\frac{x^{1-\alpha}}{1!}+\frac{x^{2-\alpha}}{2!}+\frac{x^{3-\alpha}}{3!}+\cdots$$ You see that if $\alpha \leq 2$, there is a small problem (which one ?). So, now, in principle, for the antiderivative $$\int \frac{e^x-1}{x^{\alpha}}=\sum_{i=1}^{\infty} \frac{x^{i-\alpha+1}}{i!(i-\alpha+1)}$$
I am sure that you can take from here.