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Computing integrals on the half line 
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#1
Feb1612, 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 


#2
Feb1712, 09:42 PM

P: 38

Try the punctured disc with boundary ##C_{\epsilon}\cup[\epsilon,R]\cup C_R.##



#3
Feb1712, 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.



#4
Feb1712, 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 halfplanes, 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 


#5
Feb2012, 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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