Anas.P
  • Anas.P
fourier transform question the first question please as soon as possible
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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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!
Anas.P
  • Anas.P
@mathmate
Anas.P
  • Anas.P
@e.mccormick
Anas.P
  • Anas.P
Find the fourier Transform of \[e^{-a^{2}x^{2}} ,a>0\]Given \[\int\limits_{-\infty}^{\infty} e^{-t} dt = \sqrt{\pi}\]

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Anas.P
  • Anas.P
@zzr0ck3r @Nnesha
Michele_Laino
  • Michele_Laino
hint: first step, we have to compute this integral: \[\large g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx\]
Anas.P
  • Anas.P
i tried... this method substitution method. integration by parts but it keeps rotating back with nothing reducing. if you could show me a step by step process of solving this i would really be grateful
Michele_Laino
  • Michele_Laino
now, we can write this: \[\Large \begin{gathered} {a^2}{x^2} - i\omega x = {a^2}{x^2} - i\omega x - \frac{{{\omega ^2}}}{{4{a^2}}} + \frac{{{\omega ^2}}}{{4{a^2}}} = \hfill \\ \hfill \\ = {\left( {ax - \frac{{i\omega }}{{2a}}} \right)^2} + \frac{{{\omega ^2}}}{{4{a^2}}} \hfill \\ \end{gathered} \] so we get: \[\Large \begin{gathered} g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx = \hfill \\ \hfill \\ = {e^{ - \frac{{{\omega ^2}}}{{4{a^2}}}}}\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {ax - \frac{{i\omega }}{{2a}}} \right)}^2}}}} dx \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now we have to make this variable change: \[\Large z = ax - \frac{{i\omega }}{{2a}}\]
Michele_Laino
  • Michele_Laino
where z is the new variable
Michele_Laino
  • Michele_Laino
hint: we can use this identity: \[\Large \int_{ - \infty }^{ + \infty } {{e^{ - {z^2}}}} dz = 2\int_0^{ + \infty } {{e^{ - {z^2}}}dz} \]
Anas.P
  • Anas.P
Thanks man... Really helpful... I had no idea that we had to use the complete the square method. Thanks again.

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