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Homework Help
Calculus and Beyond Homework Help
Find, using complex contour integrals, the function f(x)....
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[QUOTE="Poirot, post: 5474031, member: 587460"] [h2]Homework Statement [/h2] whose Fourier transform is f[SUP]~[/SUP](p) = 1/(a[SUP]2[/SUP] + p[SUP]2[/SUP]) [h2]Homework Equations[/h2] f(x) = 1/√2π ∫[SUP]∞[/SUP][SUB]-∞[/SUB] e[SUP]ipx[/SUP] f[SUP]~[/SUP](p)[h2]The Attempt at a Solution[/h2] First of all I let f(z) = e[SUP]ixz[/SUP]/(z[SUP]2[/SUP] + a[SUP]2[/SUP]) and γ = γ[SUB]1[/SUB] + γ[SUB]2[/SUB] with the ϒ's parametrised by: γ[SUB]1[/SUB] : {z=t, -R<t<R} γ[SUB]2[/SUB] : {z=Re[SUP]it[/SUP], 0<t<π} (So a semicircle of radius R) In this contour the only pole that lies inside is the z= +ia so using Cauchy's residue theorem: ∫[SUB]ϒ[/SUB]f(z)dz = 2πi (Res(f, z=ia)) I found the residue of z=ia to be Rez(f,z=ia)= -i e[SUP]-ax[/SUP]/2a So ∫[SUB]ϒ[/SUB]f(z)dz= π e[SUP]-ax[/SUP]/a And I have something in my notes about the fact you have to check that the integral of f(z) on γ[SUB]2[/SUB] goes to zero as R goes to infinity, which I vaguely understand because we actually want the integral from -∞ to +∞. So the checks I have in my notes are: Suppose γ has length L and on γ |f(z)|<M Then |∫[SUB]γ[/SUB]f(z)dz|≤M⋅L With γ ≡ {z=γ(t), a≤t≤b} L = ∫[SUB]a[/SUB][SUP]b[/SUP]dt|γ'(t)| On γ[SUB]2[/SUB]: |z| =R Using ||a|-|b|| ≤ |a+b| ≤ |a|+|b| R[SUP]2[/SUP]-a[SUP]2[/SUP] ≤ |z[SUP]2[/SUP]+a[SUP]2[/SUP]| ≤ R[SUP]2[/SUP]+a[SUP]2[/SUP] So 1/R[SUP]2[/SUP]-a[SUP]2[/SUP] ≥ 1|z[SUP]2[/SUP]+a[SUP]2[/SUP]| ≥ 1R[SUP]2[/SUP]+a[SUP]2[/SUP] Therefore |∫[SUB]γ[SUB]2[/SUB][/SUB]f(z)dz|≤πR/(R[SUP]2[/SUP]-a[SUP]2[/SUP]) So ∫[SUB]γ[SUB]2[/SUB][/SUB]f(z)dz → 0 as R → ∞ as you have |∫[SUB]γ[SUB]2[/SUB][/SUB] f(z)dz| ≤ e[SUP]-Rxsin(t)[/SUP]/(R[SUP]2[/SUP] - a[SUP]2[/SUP]) and if x ≥0 the exponent is, at most, equal to 1, but if x<0, it blows up and we need a different contour for x<0. Let γ = γ[SUB]1[/SUB] + γ[SUB]3[/SUB] with ϒ[SUB]3[/SUB] : {z=Re[SUP]it[/SUP], π<t<2π} so a semicircle in the negative part, and this time it's a clockwise contour so Cauchy's residue must carry a minus sign on the right hand side. So doing everything as before I found ∫[SUB]ϒ[/SUB] = πe[SUP]ax[/SUP]/a But I'm having trouble with the checks, I found that the inequality for the modulus of the integral over ϒ[SUB]3[/SUB] comes out the same? So effectively it still blows up? But I know this is the right answer I just can't follow through with the checks. I appreciate this is quite long, but a lot of it is just background to set the scene, and I think I just have a lack of understanding of the checks and just know the procedure. Thanks in advance [/QUOTE]
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Calculus and Beyond Homework Help
Find, using complex contour integrals, the function f(x)....
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