Can the Complex Integral Problem Be Solved Using Residue Theorem?

[Jenny short]

I have this problem with a complex integral and I'm having a lot of difficulty solving it:

Show that for R and U both greater than 2a, \exists C > 0, independent of R,U,k and a, such that $$\int_{L_{-R,U}\cup L_{R,U}} \lvert f(z)\rvert\,\lvert dz\rvert \leqslant \frac{C}{kR}.$$

Where a > 0, k > 0, $$L_{-R,U} = \{-R + iy \mid 0 \leqslant y \leqslant U\}$$, $$L_{R,U} = \{ R + iy \mid 0 \leqslant y \leqslant U\}$$ and $$f(z) := \frac{z e^{ikz}}{z^2+a^2}$$I don't really know where to start, or what to use. Any help would be greatly appreciated, thanks

 

[TJGilb]

I would start would by taking the antiderivative of $f(z) = \frac {ze^{ikz}} {z^2 + a^2}$ with respect to z.

 

[Svein]

I tried to suggest an answer in an earlier thread, but now the problem has changed somewhat. So, I will give the answer to a question that has not been asked (otherwise I will get a reprimand).

So: Let Γ be the curve you have specified. Now factorize the denominator in f(x): $\frac{ze^{ikz}}{z^{2}+a^{2}}=\frac{ze^{ikz}}{(z+i\cdot a)(z-i\cdot a)}$. Since R and U are both greater than 2a, the points i⋅a and (-i)⋅a both are inside Γ. Furthermore, the residue at $z=i\cdot a$ is $\frac{ia\cdot e^{-ka}}{2ia}=\frac{e^{-ka}}{2}$ and the residue at $z=-i\cdot a$ is $\frac{-ia\cdot e^{ka}}{-2ia}=\frac{e^{ka}}{2}$. Thus, the sum of the residues is $\cosh(ka)$ and the value of the integral $\int_{\Gamma}\frac{ze^{ikz}}{z^{2}+a^{2}}=2\pi i\cosh(ka)$.