for what values of the real number x is the improper integral [integral from 1 to infinity] t^(x-1)e^(-t)dt convergent?
my solution was:
1<= t < infinity
suppose x < 1
and define f(t) = t^(x-1)e^(-t) = e^(-t)/t^(1-x) , g(t) = e^(-t)
now x < 1, so x-1 < 0, which implies that 1 - x > 0
and since we are on the interval 1 <= t < infinity, 1/t^(1-x) <= 1
so f(t) <= g(t) for t >= 1, and by the comparison theorem if g(t) converges then so does f(t)
[integral from 1 to infinity] g(t) = [integral from 1 to infinity] e^(-t)dt = 1/e
so g(t), converges for all x < 1, and by the comparison theorem so does f(t)
I used a similar proof using L'hospital's rule t prove the case for x >= 1
but my question is: is this right (for x < 1)??
because say I had chosen g(t) = 1/t^(1-x)
g(t) >= f(t) for all t >= 1 because f(t) = e^(-t)/t^(1-x) and e^(-t) = 1/e^t < 1 for t >=1
so using the comparison theorem with this g(t), if g(t) converges then so does f(t)
[integral from 1 to infinity] g(t) = [integral from 1 to infinity] 1/t^(1-x) *dt
which converges if 1-x > 1 ==> x < 0,
and diverges if 1-x <= 1 ==> 0 <= x
so I'm getting two different answers!! what am I doing wrong!?!?