What Happens When Evaluating Improper Integrals with Limit?

JIHUN
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Homework Statement


Evaluating the following formula:
upload_2015-10-23_22-41-31.png

The Attempt at a Solution


Since the integral part is unknown, dividing the case into two: converging and diverging
If converging: the overall value will always be 0
If diverging: ...?
 
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As you say, if the integral converges for ##x \to 0^+##, the result is always zero. However, even if the integral diverges, it is still possible that one can get a finite value. This depends on the speed of divergence. Try it out with ##f(t) = 1## and ##f(t) = 1/t##. What happens for ##f(t) = 1/t^\alpha, \alpha > 1##?
 
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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