Prove that the following integral exists I have the following question:
Suppose that $ f(z) $ is analytic at every point of the closed domain $$ |z| ≥ R_0$$$$ 0 ≤ arg (z) ≤ α $$ where $ (0 ≤ α ≤ 2π)$, and that $$\lim_{z\to \infty} zf(z) = 0. $$ 
Prove that if the integral
$$ J_1 =\int_0^\infty f(x)dx $$
exists, then so does the integral
$$ J_2 =\int_L f(z)dz $$
where $L$ is the ray $z = re^{i\alpha}$, $0 ≤ r ≤ ∞$. Moreover, show that $J_1 = J_2$.
The hint that I was given was to use Cauchy's Theorem and the fact that under the same domain, if $$\lim_{z\to\infty} zf(z) = A,$$ then $$\lim_{R\to\infty} \int_L f(z)dz = iA\alpha, $$ where $L$ is the arc of the circle $|z| = R$ that lies in the domain in question (this was a proof I did earlier). 
My approach so far: 
Using the second hint, we can let $A = 0$ because $\lim_{z\to \infty} zf(z) = 0. $ was given to us. Therefore, we can claim that because our domain from this hint is the same as the domain in question, we can claim that
$$J_2 = \int_L f(z)dz = 0$$
So, we have seen that $J_2$ exists and that its value is $0$. This is also evident in Cauchy's Integral Theorem, which works in a simply connected domain where $f(z)$ is analytic. However, what would I do to incorporate $J_1$ into the rest of this proof? I know that showing $J_1 = J_2$ is still left, but I was wondering if there was another piece of the proof I was missing that allows me to consider it?
 A: I am assuming, per my comment, that the domain is $|z| \ge 0$, not $R_0$.
You are calling two different curves $L$ here, and have forgotten which is which. $J_2$ is integration along the curve $L_2 = \{re^{i\alpha}\mid r \ge 0\}$, so $$J_2 = \int_{L_2} f(x)dz = \int_0^\infty f(re^{i\alpha})e^{i\alpha}dr$$
Your theorem says if $L_C = \{Re^{it}\mid 0\le t\le \alpha\}$ for some $R$ and $\lim_{z \to \infty} zf(z) = A$, then $$\lim_{R\to\infty} \int_{L_C}f(z)dz = \lim_{R\to\infty} \int_0^\alpha f(Re^{it})ie^{it}dt = 0$$
So your theorem does not say that $J_2 = 0$. It's a different integral. Nor does Cauchy's theorem say anything about $J_2$ directly. Cauchy's theorem is about closed curves, and $J_2$ is obviously not closed.
What you want to consider is the curve $C = C(R)$ for each $R > 0$ that starts at $0$, follows the real line out to $R$, then proceeds counter-clockwise along the circle about $0$ of radius $R$ until it gets to the point $Re^{i\alpha}$. From there there it follows the ray $L_2$ backwards to get back to $0$. That is a closed curve, so Cauchy's theorem applies to it.
