View Single Post

## Computing integrals on the half line

Notice that the inverse Fourier transform of the Heaviside step function:
$$\int_{-\infty}^{\infty}{\frac{d k}{2\pi} \, \theta(k) \, e^{i \, k \, x}} = -\frac{1}{2\pi \, i \, x}, \ \mathrm{Im}x > 0$$
Thus, we may represent the Heaviside step function as:
$$\theta(k) = -\frac{1}{2\pi \, i} \, {d t \, \frac{e^{-i \, k \, t}{t + i \, \eta}}, \ \eta \rightarrow +0$$

Why do we need it? Because your integral goes to:
$$\int_{-\infty}^{\infty}{f(k) \, e^{i \, k \, x} \, \theta(k)}$$
If you substitute the integral representation for the step function and change the order of integration, you should get:
$$-\frac{1}{2\pi \, i} \, \int_{-\infty}^{\infty}{\frac{d t}{t + i \, \eta} \, \int_{-\infty}{\infty}{f(k) \, e^{i \, k \, (x - t)}}}$$
Now, you may use the residue theorem for the integral over k, but you need to close the contour in different half-planes, depending on whetgher $x > t$ or $x < t$. The remaining integral over t is again over the whole real line, but , due to the above conditions, should be split into $-\infty$ to x, and from x to [itex]\infty]. Then, making a sub