Domain of validity of an integral

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Homework Statement



This integral is actually part of a question I had on an exam about analytic continuation. The integral is
\int_0^1 dx \frac{x^n}{\sqrt{x^3+5}}. The first part of the question is "what is its domain of validity for absolute convergence of the integral?" It then goes on to ask for a valid analytic continuation (I know how to do that part)


Homework Equations



none


The Attempt at a Solution



The answer key my professor provided states that "the integrand only blows up at x=0, if at all, in a finite range of integration. As x -> 0, the integrand is O(x^n), so there is absolute convergence if Real n > -1.

I can easily how how this is the domain of validity assuming the integral blows up at 0, but I can't see how the integral would possibly blow up at 0. Plugging in zero for x gives \frac{0}{\sqrt{5}}. I've tried putting the integral in other forms and I still don't see it. I taylor expanded the square root around x = 0 but that still doesn't cause it to blow up, nor does multiplying the top and bottom of the integral by the square root. I'm sure I'm just making some stupid calculus mistake somewhere...
 
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tourjete said:

Homework Statement



This integral is actually part of a question I had on an exam about analytic continuation. The integral is
\int_0^1 dx \frac{x^n}{\sqrt{x^3+5}}. The first part of the question is "what is its domain of validity for absolute convergence of the integral?" It then goes on to ask for a valid analytic continuation (I know how to do that part)


Homework Equations



none


The Attempt at a Solution



The answer key my professor provided states that "the integrand only blows up at x=0, if at all, in a finite range of integration. As x -> 0, the integrand is O(x^n), so there is absolute convergence if Real n > -1.

I can easily how how this is the domain of validity assuming the integral blows up at 0, but I can't see how the integral would possibly blow up at 0. Plugging in zero for x gives \frac{0}{\sqrt{5}}. I've tried putting the integral in other forms and I still don't see it. I taylor expanded the square root around x = 0 but that still doesn't cause it to blow up, nor does multiplying the top and bottom of the integral by the square root. I'm sure I'm just making some stupid calculus mistake somewhere...

Did you not read the restriction Re(n) > -1? What happens if you take Re(n) = -1 (for example, n = -1)? What happens if you take Re(n) < -1, for example, n = -2?
 
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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