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anonymous
 one year ago
fourier transform question
the first question please
as soon as possible
anonymous
 one year ago
fourier transform question the first question please as soon as possible

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the fourier Transform of \[e^{a^{2}x^{2}} ,a>0\]Given \[\int\limits_{\infty}^{\infty} e^{t} dt = \sqrt{\pi}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: first step, we have to compute this integral: \[\large g\left( \omega \right) = \int_{  \infty }^{ + \infty } {{e^{  {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{  \infty }^{ + \infty } {{e^{  \left( {{a^2}{x^2}  i\omega x} \right)}}} dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.2now, 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
 one year ago
Best ResponseYou've already chosen the best response.2now we have to make this variable change: \[\Large z = ax  \frac{{i\omega }}{{2a}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2where z is the new variable

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: we can use this identity: \[\Large \int_{  \infty }^{ + \infty } {{e^{  {z^2}}}} dz = 2\int_0^{ + \infty } {{e^{  {z^2}}}dz} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks man... Really helpful... I had no idea that we had to use the complete the square method. Thanks again.
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