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Find Fourier cosine transform of $e^{-a^2 x^2}$ and hense evaluate Fourier sine transform of $x\cdot e^{-a^2x^2}$.

I can solve this question only if there is $x$ instead of $x^2$ in the exponential function $e^{-a^2x^2}$. Because in this situation i can use integral formula :- $\int e^{-ax} \cos sx\,dx = \frac{e^{-ax}}{a^2+x^2} \left( s\sin sx-a\cos sx \right)$ but what should i do if there is $x^2$ in exponential function $e$??

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    Please Learn TeX it will be useful.2010-12-16

2 Answers 2

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HINT 1:

$ \int_{ - \infty }^\infty {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x} = \int_{ - \infty }^\infty {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x}. $

HINT 2:

Characteristic function of the (centered) normal distribution: see this.

EDIT: The first hints didn't help. So,

$ \int_{ - \infty }^\infty {\cos (sx)e^{ - a^2 x^2 } \,{\rm d}x} = \int_{ - \infty }^\infty {[\cos (sx) + {\rm i}\sin (sx)]e^{ - a^2 x^2 } \,{\rm d}x} = \int_{ - \infty }^\infty {e^{{\rm i}sx} e^{ - a^2 x^2 } {\rm d}x} = ? $ In order to calculate the right-most integral, bring it to the form(s) b \int_{ - \infty }^\infty {e^{ - a^2(x - {\rm i}c )^2 } \, {\rm d}x} = b \int_{ - \infty }^\infty {e^{ - a^2 x^2 } \,{\rm d}x} = b' \int_{ - \infty }^\infty {e^{ - x^2 } \,{\rm d}x} =...

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    @Lokesh: http://www.wolfr$a$m$a$lph$a$.com/. (The result involves error functions.)2010-12-16
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Find the Fourier transform of $e^{-a^2x^2}$. The real part of that is relevant to your cosine transform.

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    Of course, I realized that.2010-12-16