- #1
sikrut
- 49
- 1
Using contour integration and the residue theorem, evaluate the following
"Fourier" integral:
[tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
with real-valued constants [itex]\Gamma > 0[/itex] and [itex]\Omega[/itex]. Express your answers in terms of [itex]t, \Gamma[/itex] and [itex]\Omega[/itex].
Hints: Before starting the contour integral formulation:
(1) Since [itex]cos(\omega t) = Re(e^{i \omega t})[/itex] and [itex]sin(\omega t) = Im(e^{i \omega t})[/itex], write F1 as either the real part or the imaginary part of complex-valued integrals where the cos- and sin-functions have been replaced by the eiwt function, using e.g., [itex] \int_{-\infty}^{+\infty} Re[f(\omega)]d\omega = Re[\int_{-\infty}^{+\infty} f(\omega)d\omega][/itex].
(2) Get rid of the [itex]\Gamma[/itex] and [itex]\Omega[/itex] parameters insde the integrals, by the following substitutions:
express [itex]\omega[/itex] by [itex]x := (\omega - \Omega)/\Gamma;[/itex]
[itex]t[/itex] by [itex]s := \Gamma t,[/itex]
and [itex]\Omega[/itex] by [itex]\nu := \Omega/\Gamma.[/itex]
(3) Depending on the sign of t, then "close" the real-axis contour with a semi-circle of radius R either in the upper half or in the lower half of the complex plane such that contribution from the semi-circle vanishes in the limit of [itex]R \rightarrow \infty[/itex].
(5) Use your Residue Recipes to find the residues and complete the calculation.
Attempt:
[tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
[tex]= \int_{-\infty}^\infty \frac{\Gamma Im(e^{i \omega t})}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
[tex]= Im(\int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw)[/tex]
[itex]\omega \rightarrow x:= (\omega - \Omega)/\Gamma \rightarrow \omega = \Gamma x + \Omega \rightarrow d\omega = \Gamma dx[/itex]
[itex]t \rightarrow s := \Gamma t \rightarrow t= s/\Gamma [/itex]
[itex]\Omega \rightarrow \nu := \Omega/\Gamma[/itex]
[itex]\omega t = \frac{\Gamma x + \Omega)s}{\Gamma} = (x + \frac{\Omega}{\Gamma})s = (x+\nu)s[/itex]
where: [itex] I = \int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw[/itex]
[tex] I = \int_{-\infty}^\infty \frac{\Gamma^2 e^{i(x+\nu)s}}{\Gamma^2 (x^2 + 1)}dx = \int_{-\infty}^\infty \frac{e^{i(x+\nu)s}}{(x^2 + 1)}dx = e^{i \nu s}\int_{-\infty}^\infty \frac{e^{ixs}}{(x+i)(x-i)}dx[/tex]Now, I know that the next step is concatenating the line with with the semicircle, but I am not quite sure how to go about setting it up:
[tex] I = \int_{I_R^+} + \int_{C_R^+} - \int_{C_R^-}[/tex]
[tex] I = \oint_{P_R^+} -\int_{C_R^-}[/tex]
"Fourier" integral:
[tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
with real-valued constants [itex]\Gamma > 0[/itex] and [itex]\Omega[/itex]. Express your answers in terms of [itex]t, \Gamma[/itex] and [itex]\Omega[/itex].
Hints: Before starting the contour integral formulation:
(1) Since [itex]cos(\omega t) = Re(e^{i \omega t})[/itex] and [itex]sin(\omega t) = Im(e^{i \omega t})[/itex], write F1 as either the real part or the imaginary part of complex-valued integrals where the cos- and sin-functions have been replaced by the eiwt function, using e.g., [itex] \int_{-\infty}^{+\infty} Re[f(\omega)]d\omega = Re[\int_{-\infty}^{+\infty} f(\omega)d\omega][/itex].
(2) Get rid of the [itex]\Gamma[/itex] and [itex]\Omega[/itex] parameters insde the integrals, by the following substitutions:
express [itex]\omega[/itex] by [itex]x := (\omega - \Omega)/\Gamma;[/itex]
[itex]t[/itex] by [itex]s := \Gamma t,[/itex]
and [itex]\Omega[/itex] by [itex]\nu := \Omega/\Gamma.[/itex]
(3) Depending on the sign of t, then "close" the real-axis contour with a semi-circle of radius R either in the upper half or in the lower half of the complex plane such that contribution from the semi-circle vanishes in the limit of [itex]R \rightarrow \infty[/itex].
(5) Use your Residue Recipes to find the residues and complete the calculation.
Attempt:
[tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
[tex]= \int_{-\infty}^\infty \frac{\Gamma Im(e^{i \omega t})}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
[tex]= Im(\int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw)[/tex]
[itex]\omega \rightarrow x:= (\omega - \Omega)/\Gamma \rightarrow \omega = \Gamma x + \Omega \rightarrow d\omega = \Gamma dx[/itex]
[itex]t \rightarrow s := \Gamma t \rightarrow t= s/\Gamma [/itex]
[itex]\Omega \rightarrow \nu := \Omega/\Gamma[/itex]
[itex]\omega t = \frac{\Gamma x + \Omega)s}{\Gamma} = (x + \frac{\Omega}{\Gamma})s = (x+\nu)s[/itex]
where: [itex] I = \int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw[/itex]
[tex] I = \int_{-\infty}^\infty \frac{\Gamma^2 e^{i(x+\nu)s}}{\Gamma^2 (x^2 + 1)}dx = \int_{-\infty}^\infty \frac{e^{i(x+\nu)s}}{(x^2 + 1)}dx = e^{i \nu s}\int_{-\infty}^\infty \frac{e^{ixs}}{(x+i)(x-i)}dx[/tex]Now, I know that the next step is concatenating the line with with the semicircle, but I am not quite sure how to go about setting it up:
[tex] I = \int_{I_R^+} + \int_{C_R^+} - \int_{C_R^-}[/tex]
[tex] I = \oint_{P_R^+} -\int_{C_R^-}[/tex]
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