for what values of the real number x, is the [integral from 0 to 1] t^(x-1) * e^(-t) dt improper? For which of these values is the improper integral convergent?
This integral has a discontinuity at t=0 if x - 1 < 0, because making the exponent positive would require us to rewrite t^(x-1) as 1/t^(1-x) where 1 - x > 0, and we would divide by a negative if t = 0
so the integral is improper for x < 1
What is a concise way to figure out where it converges? I have a very long, very messy solution to arrive at it converges if x > 0 (which i could type up if necessary), but I'm sure there is a better way. Can someone show me?
This integral has a discontinuity at t=0 if x - 1 < 0, because making the exponent positive would require us to rewrite t^(x-1) as 1/t^(1-x) where 1 - x > 0, and we would divide by a negative if t = 0
so the integral is improper for x < 1
What is a concise way to figure out where it converges? I have a very long, very messy solution to arrive at it converges if x > 0 (which i could type up if necessary), but I'm sure there is a better way. Can someone show me?