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.
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.