On the orientation of the chosen path for evaluating improper integrals by the Residue's Theorem.

Question

I'm studying some lecture notes on field theory and eventually an integral like this needs to be evaluated

$$ \int_{-\infty}^{\infty} dx \; e^{-ixt}f(x) $$

Proceeding by countour integration, the author of the notes states that 'for $t>0$ we move the integration contour from the real axis to a distance $a$ into the lower half of the complex plane', and says later that 'obviously we can apply the same argument for negative $t$, in that case we shift the integration contour into the upper complex plane'.

My question is not about the calculations that need to follow to solve the problem, which I understand, but rather why does the coefficient of the $ix$ on the argument of the exponential determines the orientation that I must choose for the path? I understand that not knowing this is likely to mean that I don't fully understand the method for evaluating this kind of integrals, nevertheless I exhaustively searched for what I'm missing and couldn't find it.

Answer 1

We want the integrand to vanish over that part of the contour. In other words, we want $\Re(-ixt) < 0$. For $t > 0$, it means $\Re(ix) > 0$, or $\Re(x) < 0$, so we chose the lower half.