Convergence test of the following improper integral $\int_0^\infty \frac {e^{-1/x}-1} {\ x^{2/3}}dx$

Question

I've been trying for a couple of hours to prove the convergence of the following integral: $$\int_0^\infty \frac {e^{-1/x}-1} {\ x^{2/3}}dx$$ Eventually I understood from Wolfram-Alpha that the integral converges to $\Gamma(-1/3)$ and it makes sense because by substitution $t=1/x $ I get the following integral: $$\int_0^\infty \frac {e^{-t}-1} {\ t^{4/3}}dt$$ And it's pretty close to $\Gamma(-1/3)$ except for the $-1$ that bothers me. What have I done wrong until this point? And how can I prove it? Thank you.

Answer 1

Hint: Let $I(k)=\displaystyle\int_0^\infty\frac{e^{-kx}-1}{x^n}~dx$. Now, evaluate $I'(k)$ by differentiating under the integral sign. Then integrate that expression, and let $k=1$.