Closing Contours 'Up' and 'Down' in a Contour Integral

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I am considering an integral which can be evaluated using techniques from complex variables. The first step is to simplify the integral.

$$\int^{\infty}_{-\infty}dk\:\frac{e^{ikr}-e^{-ikr}}{k} = \lim_{\delta \rightarrow 0} \Bigg[ \int^{\infty}_{-\infty}dk\:\frac{e^{ikr}-e^{-ikr}}{k+i\delta} \Bigg]$$

Obviously if $\delta >0$, then there is a pole at $-i \delta$. However, the author then says that for the $e^{ikr}$ part, we close the contour up in the contour integral to get exponential decay for large $k$: the contour then misses the pole to get a zero result. For the $e^{-ikr}$ part we have to close the contour down such that there is a loop around the pole which is evaluated using the residue theorem to get a final non-zero result.

It might just be the 'idiomatic' writing style of the author but I am just not sure what is meant when the author talks about closing up and closing down in this context: could someone clarify for me and maybe draw the contour which is being described?

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Split the integral in two: $$ \int^{\infty}_{-\infty}dk\:\frac{e^{ikr}-e^{-ikr}}{k+i\delta} = \int^{\infty}_{-\infty}dk\:\frac{e^{ikr}}{k+i\delta} - \int^{\infty}_{-\infty}dk\:\frac{e^{-ikr}}{k+i\delta} $$

For the first integral we take the following contour:

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For the second integral we instead let the semicircle be in the negative imaginary halfplane:

enter image description here