Showing that the $\Gamma(z)$ improper integral diverges for $\operatorname{Re} z \leq 0$ I was able to show that the improper integral, $\int_{0}^{\infty} t^{z-1}e^{-t}\,dt$, converges absolutely for $\operatorname{Re} z > 0$ and that it diverges for purely real values $x \leq 0$. However, I have not been able to show that it diverges for the general complex number with $\operatorname{Re} z \leq 0$.
Letting $z = x + iy$, I thought I could show that the real part of the integral on $[0,1]$, which is $$\int_{0}^{1} t^{x-1}e^{-t}\cos(y\log t)\,dt$$ diverges when $x \leq 0$ by limit comparison with some function. We can eliminate the $e^{-t}$ like this, but the rest of the integrand is highly oscillatory near zero and evades my best attempts at a comparison.
I hope I'm going about this the right way, but it seems like there should be an easier way to show it than this.
 A: I think I solved it now. Let $z = x + i y$ with $x \leq 0$. Since divergence is known for $y = 0$ and the integral is symmetric with respect to $y$, it's enough to show that the integral diverges for $y > 0$. I'll show that the integral is not Cauchy near the lower limit.
Let $\varepsilon = \frac{\pi}{2e\sqrt{3}y}$ and let $\delta > 0$ be given. Without loss of generality, we can assume $\delta < 1$. Choose $N$ large enough so that 
$$\theta_{2} = \frac{\pi}{6} - \pi N < y\log\delta$$
Let $\theta_{1} = -\frac{\pi}{6} - \pi N$, $\varepsilon_{1} = e^{\theta_{1}/y}$, and $\varepsilon_{2} = e^{\theta_{2}/y}$, so that
$$0 < \varepsilon_{1} < \varepsilon_{2} < \delta$$
We can also choose an even larger $N'$, if necessary, so that $e^{x\theta_{2}/y} \geq 1$. By the Mean Value Theorem for integrals, we have
$$\left\lvert\int_{\varepsilon_{1}}^{\varepsilon_{2}} t^{x-1} e^{-t} \cos(y \log t)\,dt\right\rvert = \lvert\cos(y \log c)\rvert \int_{\varepsilon_{1}}^{\varepsilon_{2}} t^{x-1} e^{-t}\,dt$$
for some $c \in (\varepsilon_{1}, \varepsilon_{2})$. Then $y\log c \in (\theta_{1},\theta_{2})$ so that 
$$\lvert\cos(y \log c)\rvert \geq \frac{\sqrt{3}}{2}$$
After replacing $e^{-t}$ with $e^{-1}$ and substituting $u = \log(t)$ we have that
$$\lvert\cos(y \log c)\rvert \int_{\varepsilon_{1}}^{\varepsilon_{2}} t^{x-1} e^{-t}\,dt \geq \frac{\sqrt{3}}{2e} \int_{\theta_{1}/y}^{\theta_{2}/y} e^{xu}\,du$$
By our choice of $N'$, the integral will be larger than $\frac{\theta_{2}}{y} -
\frac{\theta_{1}}{y} = \frac{\pi}{3y}$. Hence
$$\frac{\sqrt{3}}{2e} \int_{\theta_{1}/y}^{\theta_{2}/y} e^{xu}\,du \geq \frac{\pi}{2e\sqrt{3}y} = \varepsilon$$
Since the integral fails the Cauchy criterion, it diverges for $x \leq 0$ and $y > 0$, and so it diverges for $x \leq 0$ by symmetry.
