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Dickfore
Dickfore is offline
#4
Feb17-12, 10:25 PM
P: 3,015

Computing integrals on the half line


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

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