Question

fourier transform question
the first question please
as soon as possible

Answer 1

@mathmate

Answer 2

@e.mccormick

Answer 3

Find the fourier Transform of \[e^{-a^{2}x^{2}} ,a>0\]Given \[\int\limits_{-\infty}^{\infty} e^{-t} dt = \sqrt{\pi}\]

Answer 4

@zzr0ck3r
@Nnesha

Answer 5

hint:
first step, we have to compute this integral:
\[\large g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx\]

Answer 6

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

Answer 7

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:
\[\Large \begin{gathered}
g\left( \omega \right) = \int_{ - \infty }^{ + \infty } {{e^{ - {a^2}{x^2}}}{e^{i\omega x}}} dx = \int_{ - \infty }^{ + \infty } {{e^{ - \left( {{a^2}{x^2} - i\omega x} \right)}}} dx = \hfill \\
\hfill \\
= {e^{ - \frac{{{\omega ^2}}}{{4{a^2}}}}}\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {ax - \frac{{i\omega }}{{2a}}} \right)}^2}}}} dx \hfill \\
\end{gathered} \]

Answer 8

now we have to make this variable change:
\[\Large z = ax - \frac{{i\omega }}{{2a}}\]

Answer 9

where z is the new variable

Answer 10

hint:
we can use this identity:
\[\Large \int_{ - \infty }^{ + \infty } {{e^{ - {z^2}}}} dz = 2\int_0^{ + \infty } {{e^{ - {z^2}}}dz} \]

Answer 11

Thanks man... Really helpful... I had no idea that we had to use the complete the square method.
Thanks again.