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For the integral I = 1/(1+x)^e how do we known if it converges or diverges. Upper Limit of 0 and lower limit of -1.

I know that it is improper - is it unbounded at x=-1?

My understanding is that this integral diverges as N approaches -1 from above. Does the function become undefined?

Thanks

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  • $\begingroup$ What is $e$? Is it the base of the natural logarithm, or some arbitrary real constant? $\endgroup$ – Daniel Fischer May 7 '15 at 0:00
  • $\begingroup$ natural logarithm $\endgroup$ – Oscar John May 7 '15 at 0:02
  • $\begingroup$ Try applying the power rule to the integral going from $a$ to $0$ and then send $a \to 0^+$. You should find that the integral indeed diverges (to $+\infty$, specifically). $\endgroup$ – Ian May 7 '15 at 0:03
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$$\int_{-1}^0 {dx\over (1 + x)^e} = \int_0^1{dx\over x^e}$$ What happens at 0?

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