Rose16
  • Rose16
find the Fourier transform for 1/(a^2 +x^2 ) in 2 ways 1- by complex integration if you can, if not, look it up in a table of integrals. 2- by using the result of the Fourier transform for exp-cx.
Physics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Rose16
  • Rose16
I took a picture to my question that I posted 30 mins ago to make it clearer
IrishBoy123
  • IrishBoy123
\[ \mathcal{F} \{ e^{-c|x|} \} =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-c|x|}e^{-i\omega x}\, dx\] \[\sqrt{2 \pi} \mathcal{F} \{ e^{-c|x|} \}= \int_{-\infty}^{0}e^{cx}e^{-i\omega x}\, dx+\int_{0}^{\infty}e^{-cx}e^{-i\omega x}\, dx \] \[= \int_{-\infty}^{0}e^{cx-i\omega x}\, dx+\int_{0}^{\infty}e^{-cx-i\omega x}\, dx \] \[= \left[ \frac{e^{(c-i\omega)x}}{c-i\omega} \right]_{-\infty}^0-\left[ \frac{e^{-(c+i\omega)x}}{c+i\omega} \right]_{0}^{\infty}\] \[=\frac{1}{c-i\omega}+\frac{1}{c+i\omega} \] \[ \mathcal{F} \{ e^{-c|x|} \} =\frac{1}{\sqrt{2 \pi}}. \frac{2c}{c^2+\omega^2}\] i think the following are the steps the question really requires it follows from the reverse transform that \[ e^{-c|x|} =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \frac{2c}{c^2+\omega^2}e^{i\omega x}\, d\omega \] sub -x for x \[ e^{-c|x|} =\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{2c}{c^2+\omega^2}e^{-i\omega x}\, d\omega \ \] switch x and \(\omega\) \[ e^{-c|\omega|} =\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{2c}{c^2+x^2}e^{-i\omega x}\, dx \ \] \[ \therefore e^{-c|\omega|} \frac{\sqrt{2 \pi} } {2c} =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{1}{c^2+x^2}e^{-i\omega x}\, dx \ = \mathcal{F} \{ \frac{1}{c^2+x^2} \}\]
IrishBoy123
  • IrishBoy123
you also want complex integration. if you mean contour integration, that is on my to do list but @Michele_Laino is a dab hand you can also do this using polar substitution and combining that with the complex exponential but that might not be what you mean

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Michele_Laino
  • Michele_Laino
going to the complex plane z, we have the subsequent formula for the Fourier transform: \[\Large \hat f\left( \omega \right) = \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} \] the integramd function has two poles, namely: \[\Large {z_1} = ia,\;{z_2} = - ia\] so, in order to evaluate that integral, we have to apply the Jordan Lemma, and we get this: \[\Large \begin{gathered} \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} = 2\pi i\left( {{{\left. {{\text{Res}}\;F} \right|}_{z = ia}}} \right),\;\;{\text{if }}\omega > 0 \hfill \\ \hfill \\ \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} = - 2\pi i\left( {{{\left. {{\text{Res}}\;F} \right|}_{z = - ia}}} \right),\;\;{\text{if }}\omega < 0 \hfill \\ \end{gathered} \] where: \[\Large F\left( z \right) = \frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}\] and the subsequent integral, evaluated along a little circumference, whose radius goes to zero, ant centered at point z=z0: \[\Large {\left. {{\text{Res}}\;F} \right|_{{z_0}}} = \frac{1}{{2\pi i}}\oint {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} \] is the residual value of F(z) at point z0. Now performing all computations, we get: \[\Large \begin{gathered} \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} = 2\pi i\left( {{{\left. {{\text{Res}}\;F} \right|}_{z = ia}}} \right) = \frac{{{e^{ - \omega a}}}}{{2a}},\;\;{\text{if }}\omega > 0 \hfill \\ \hfill \\ \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} = - 2\pi i\left( {{{\left. {{\text{Res}}\;F} \right|}_{z = - ia}}} \right) = \frac{{{e^{\omega a}}}}{{2a}},\;\;{\text{if }}\omega < 0 \hfill \\ \end{gathered} \] Such resuult can be written in a more compact form, as below: \[\Large \hat f\left( \omega \right) = \int_{ - \infty }^{ + \infty } {\frac{{{e^{i\omega z}}}}{{{z^2} + {a^2}}}} = \frac{{{e^{ - \left| \omega \right|a}}}}{{2a}}\]
Michele_Laino
  • Michele_Laino
integrand*
Michele_Laino
  • Michele_Laino
oops.. I forgot the quantity \( \Large dz \) in all integrals
IrishBoy123
  • IrishBoy123
touch of class! @Michele_Laino
Michele_Laino
  • Michele_Laino
thanks!! :) :) @IrishBoy123

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