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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. –  picakhu Dec 16 '10 at 17:02

2 Answers 2

up vote 2 down vote accepted

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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thanks for reply –  Lokesh Kumar Dec 16 '10 at 18:06
    
but, what should i do with real part ∫e^(-a²x²).cos(sx)dx= ?? –  Lokesh Kumar Dec 16 '10 at 18:13
    
i mean there is standard formula for ⌠e^(-ax)cos(sx)dx = e^(-ax)/a²+x²[bsin(sx)-acos(sx)] but what?? for this integral ∫e^(-a²x²).cos(sx)dx= ?? –  Lokesh Kumar Dec 16 '10 at 18:16
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@Lokesh: wolframalpha.com. (The result involves error functions.) –  KennyTM Dec 16 '10 at 19:33
    
thanks!! a lot.. –  Lokesh Kumar Dec 19 '10 at 11:20

Find the Fourier transform of $e^{-a^2x^2}$. The real part of that is relevant to your cosine transform.

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@Shai. While I was composing, I did not see your response. –  TCL Dec 16 '10 at 17:47
    
Of course, I realized that. –  Shai Covo Dec 16 '10 at 17:48

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