Use Cauchy's Theorem to show that if $\int_{0}^{\infty}f(x)dx$ exists, then so does $\int_{L}f(z)dz$ Suppose that $f(z)$ is analytic at every point of the closed domain $0 \leq arg z \leq \alpha$ $(0 \leq \alpha \leq 2 \pi)$, and that $\lim_{z \to \infty}z f(z) = 0$. I need to prove that if the integral $\displaystyle J_{1}=\int_{0}^{\infty}f(x) dx$ exists, then the integral $\displaystyle J_{2}=\int_{L}f(z)dz$, where $L$ is the ray $z=r e^{i \alpha}$, $0 \leq r \leq \infty$. Moreover, I need to show that $J_{1} = J_{2}$
I have been given the hint to use Cauchy's Theorem (not the Cauchy integral formula or residues - answers using either of those things are useless to me), and the result of the previous problem, which states as follows:

If $f(z)$ is continuous in the closed domain $|z|\geq R_{0}$, $0 \leq arg z \leq \alpha$ $(0 \leq \alpha \leq 2 \pi)$, and if the limit $\displaystyle \lim_{z \to \infty} zf(z) = A$ exists, then $\displaystyle \lim_{R \to \infty}\int_{\displaystyle \Gamma_{R}}f(z)dz = i A \alpha$, where $\Gamma_{R}$ is the arc of the circle $|z|=R$ lying in the given domain.

So, for this problem, I can use the fact that $\lim_{z \to \infty}zf(z) = 0$ to show that $\displaystyle \lim_{r \to \infty}\int_{\displaystyle \Gamma_{r}}f(z)dz = 0$ at some point, I guess.
Thus far, I've tried approaching this problem in two different ways.
The first way was to start out with $J_{2} = \int_{L}f(z)dz$ and then try to get $L_{1}$ to pop out somewhere. Didn't get too far with that, and anyway, I'm not sure that it is correct to write $\int_{L}f(z)dz = \lim_{r \to \infty}\int_{0}^{2\pi}f(re^{i\alpha})ire^{i \alpha}d \alpha$. All of these angles and args are confusing me, and I'm not even entirely sure what the domain on which $f(z)$ is analytic looks like.
The second way was to start out with $J_{1} = \int_{0}^{\infty}f(x) dx$, and try to parametrize it in terms of $z = re^{i \alpha}$. But, I'm not sure exactly how to do this (again, the domain is confusing. Tried to draw it; didn't help. Maybe I'm just not visualizing it right). Then, at some point, I assume I can apply Cauchy's Theorem and the given limit.
I'm guessing that since Cauchy's Theorem is involved and that the given limit goes to $0$, I'm probably going to wind up with $0 = J_{1} = J_{2}$, but I need a lot of help and guidance to show this. 
I'm at my wits end, don't have a lot of time to figure this out, and am starting to panic. Please help. 
 A: I drew a picture:

The yellow line is the part of $J_2$ of length $r$, the green line the part of $J_1$ of length $r$ and the blue line is the circle arc connecting the two lines. Note, since $f$ is analytic on the entire domain enclosed by the path and that the path is contractible in this domain, that the integral along it (going first along the yellow, then the blue, then the green or the other way around) is zero. This is the Cauchy theorem.
If we name these paths $J^1_r$, $J^2_r$ and $\Gamma_r$ (where the orientation of $\Gamma_r$ is that it goes from "top to bottom"), then the above text is just the equation:
$$\int_{J^1_r} f(z)dz = \int_{J^2_r} f(z) dz - \int_{\Gamma_r} f(z) dz$$
Your previous result is that $\lim_{r \to \infty} \int_{\Gamma_r} f(z) dz =0$ (since $\lim_{z \to \infty} z f(z) = 0$),  so taking the limit on both sides gives you
$$\int_{J_2} f(z) dz = \lim_{r \to \infty} \int_{J_2^r} f(z) dz = \lim_{r \to \infty}\left( \int_{J_1^r} f(z) dz -\int_{\Gamma_r} f(z) dz\right)  = \int_{J_1^r} f(z) dz+0$$
This shows that the integral over $J_2$ exists if the integral over $J_1$ exists, and that they are equal, which is the result you were looking for.
