# Thread: trouble with evaluating convergence of an integral

1. ## trouble with evaluating convergence of an integral

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!?!?

2. Originally Posted by minivan15
for what values of the real number x is the improper integral [integral from 1 to infinity] t^(x-1)e^(-t)dt convergent?
Notice that $t^{x-1} \leq e^{t/2}$ for $t\geq T$ for some $T\geq 1$ because exponentials overtake powers.
If makes no difference what $x$ is.

Therefore, $\int_1^{\infty} t^{x-1} e^{-t} dt = \int_1^T t^{x-1}e^{-t}dt + \int_T^{\infty}t^{x-1}e^{-t/2}dt$ (if and only if the integral converges).

Now,
$\int_T^{\infty} t^{x-1} e^{-t} dt \leq \int_T^{\infty} e^{-t/2} dt < \infty$

3. okay, I understand this!

but can you tell why my method contradicts itself? I'm okay with your answer I'd just like to see where I went wrong with my method.