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Computing integrals on the half line

by hunt_mat
Tags: computing, integrals, line
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hunt_mat
#1
Feb16-12, 05:00 PM
HW Helper
P: 1,583
Hi,

In my fluids work I have come to integrals of the type:
[tex]
\int_{0}^{\infty}\frac{e^{ikx}}{ak^{2}+bk+c}dk
[/tex]
I was thinking of evaluating this via residue calculus but I can't think of the right contour, any suggestions?

Mat
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Some Pig
#2
Feb17-12, 09:42 PM
P: 38
Try the punctured disc with boundary ##C_{\epsilon}\cup[\epsilon,R]\cup C_R.##
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pdisc.jpg  
Dickfore
#3
Feb17-12, 09:56 PM
P: 3,014
Notice that [itex]\vert e^{i \, k \, x} \vert = e^{-x \, \mathrm{Im}k}[/itex]. This means that the integral would diverge when we take the circle at infinity on the lower (upper) semicircle for positive (negative) x.

Dickfore
#4
Feb17-12, 10:25 PM
P: 3,014
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
hunt_mat
#5
Feb20-12, 04:41 AM
HW Helper
P: 1,583
I should point out that [itex]x\in\mathbb{R}[/itex]

Dick, can you explain the substitution again, I don't quite get what you're doing here and you still haven't mentioned the contour you're integrating over.


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