Finding a power series expansion for a definite integral

Szichedelic
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Homework Statement



Find a power series expansion about x = 0 for the function

f(x) = ^{1}_{0}\int\frac{1 - e^{-sx}}{s} ds


Homework Equations



The power series expansion for a function comes of the form f(x) = ^{\infty}_{0}\sum a_{k}x^{k}


The Attempt at a Solution



I've tried several things to start off with but quickly end up hitting a dead end road. First, I tried just simply taking the integral, but quickly found it isn't defined at 0 (hence why they are asking me to find a power series expansion for it). Then, I tried finding a power series expansion for the innerpart of the integral and ran into the same problem.
 
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Szichedelic said:

Homework Statement



Find a power series expansion about x = 0 for the function

f(x) = ^{1}_{0}\int\frac{1 - e^{-sx}}{s} ds

Homework Equations



The power series expansion for a function comes of the form f(x) = ^{\infty}_{0}\sum a_{k}x^{k}

The Attempt at a Solution



I've tried several things to start off with but quickly end up hitting a dead end road. First, I tried just simply taking the integral, but quickly found it isn't defined at 0 (hence why they are asking me to find a power series expansion for it). Then, I tried finding a power series expansion for the innerpart of the integral and ran into the same problem.

Did you try substituting the Taylor series (as a function of x) for ##e^{-sx}## in the integrand, simplifying and integrating?
 
Yeah, I figured that out shortly after I posted this. Can't believe I overlooked that! Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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