# For which values is this improper integral convergent?

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?

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.

• Sorry, what is the problem? And how should I continue...? – inggumnator Aug 25 '14 at 6:01
• Plug $\alpha$ equal to any number $\ge 2$ in the summation and integrate term by term and use the bounds. – Claude Leibovici Aug 25 '14 at 6:04
• You mean plug $\alpha$ into the integral, or the summation? Sorry can you explain more fully, like I tried working it out but I can't get it... I don't quite get what you're trying to say. – inggumnator Aug 27 '14 at 2:28
• Forget the last formula and consider the first term in the previous which is the Taylor expansion of the integrand. If $\alpha=1$, it is $1$ and no problem for the integral. If $\alpha=2$, it is $\frac{1}{x}$ and the antiderivative will be $\log(x)$ and so, for the bounds, the integral cannot be computed. If $\alpha=3$, it is $\frac{1}{x^2}$ and the antiderivative will be $\frac{1}{x}$ and so, for the bounds, the integral cannot be computed. – Claude Leibovici Aug 27 '14 at 3:37

Hint: near zero the integrand behaves as

$$\frac{e^x-1}{x^\alpha}\sim \frac{x}{x^\alpha}.$$