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 $e$??
Best Answer
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) $$ b \int_{ - \infty }^\infty {e^{ - a^2(x - {\rm i}c )^2 } \, {\rm d}x} = b \int_{ - \infty }^\infty {e^{ - a^2 x^2 } \,{\rm d}x} = b' \int_{ - \infty }^\infty {e^{ - x^2 } \,{\rm d}x} =... $$