Does Integration along the imaginary axis in the complex plane give an imaginary result?

Question

Given is the following contour integral :

The “arc-length” path vanishes according to Jordan’s Lemma.

Now, is it correct to use the residue theorem, and set the Integral along the imaginary axis equal to the imaginary value given by the residue theorem, and set the integral along the real axis equal to the real value ?

I’ve searched on the internet but couldn’t find a single page where this is stated, maybe it’s trivial but I just wanted to be sure.

Thanks for your help !

Answer 1

No, this is not true in general.

By definition we have $$\int_{\gamma_3} f(z) \,\mathrm{d}z = \int_0^1 f(iR(1-t)) \cdot (-iR) \,\mathrm{d}t = -iR \int_0^1 f(iR(1-t)) \,\mathrm{d}t,$$ which is imaginary for all $R \gt 0$ if and only if $f(\{z \in \mathbb C: \Re z = 0, \Im z \ge 0\}) \subset \mathbb R$ – which is false for a lot of functions.

Answer 2

You must use

$$\int_{\gamma_1}f(z)\,dz+0+\int_{\gamma_3}f(z)\,dz=\sum\text{Residues}$$

whatever the integrand is.