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## Homework Statement

I have to evaluate the following integral by means of the Cauchy Integral Theorem:

[tex]\int_{- \infty}^{\infty}\frac{e^{-ikx}}{(x+i)(x+2i)}dx[/tex]

## Homework Equations

[tex]f(z_0)=\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-z_0}dz[/tex]

## The Attempt at a Solution

The idea I had was to consider the complex contour integral,

[tex]\oint_C\frac{e^{-ikz}}{(z+i)(z+2i)}dz= \oint_C f(z)dz[/tex]

If the contour C is a semi circular contour in the lower plane with radius > 2 (so that both poles are enclosed), we can break this integral up into two parts:

[tex]\oint_C f(z)dz = \int_a^{-a}f(z)dz + \int_{arc}f(z)dz[/tex]

[tex]\implies \int_{-a}^a f(z)dz = \int_{arc}f(z)dz - \oint_C f(z)dz[/tex]

At this point I can evaluate the contour integral by splitting into two integrals using partial fractions and then directly applying Cauchy's Integral Theorem. My problem is with this arc integral, which I think goes to zero in the limit [itex]a,R \rightarrow \infty[/itex]? I know Jordan's Lemma States, for a semi circular contour in the upper plane:

[tex]\lim_{R \rightarrow \infty} \int_{C_R} e^{i \alpha z} f(z) dz = 0 [/tex]

but this formula is for alpha greater than or equal to zero. In this case, alpha is less than zero and the contour is in the lower plane.

So, does this arc term vanish or need to be evaluated, and in either case how can I proceed?