# fourier transform of f(x)=e^(-ax^2) sin(bx)

• Oct 5th 2011, 05:50 AM
wik_chick88
fourier transform of f(x)=e^(-ax^2) sin(bx)
i need to find the fourier transform of $f(x)=e^{-ax^2} sin(bx)$

i get $F(\lambda) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} sin(bx) e^{i \lambda x} dx$ (by definition)

now $sin(bx) = \frac{e^{ibx} - e^{-ibx}}{2i}$
so $F(\lambda) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} $\frac{e^{ibx} - e^{-ibx}}{2i}$ e^{i \lambda x} dx$
$= \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \int^{\infty}_{-\infty} e^{-ax^2 + ibx + i\lambda x} - e^{-ax^2 - ibx + i\lambda x} dx$

im stuck as to where to go from here! judging by our notes the next step should be
$= \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \sqrt{\frac{\pi}{a}} \int^{\infty}_{-\infty} e^{-\frac{(b+\lambda)^2}{4a}} - e^{-\frac{(b-\lambda)^2}{4a}}$
but i have no idea why this should be (and not even sure its the right step!)
• Oct 5th 2011, 06:49 AM
girdav
Re: fourier transform of f(x)=e^(-ax^2) sin(bx)
You will be able to conclude if you know the value of $I(a,b):=\int_{-\infty}^{+\infty}e^{-ax^2}e^{ibx}dx$. If you fix $a$, differentiate this with respect to $b$, and integrate by parts, you will find a differential equation.
• Oct 9th 2011, 06:12 AM
wik_chick88
Re: fourier transform of f(x)=e^(-ax^2) sin(bx)
im not sure what you mean here.why would i differentiate it with respect to b?
• Oct 9th 2011, 06:22 AM
girdav
Re: fourier transform of f(x)=e^(-ax^2) sin(bx)
Fix $a$ and put $G(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}dx$. We have $G'(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}ixdx=i\left[-e^{-ax^2}\frac 2ae^{itx}\right]_{-\infty}^{+\infty}+\frac{2i}a\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}itdx$ hence $G'(t)=-\frac{2t}aG(t)$.
• Oct 9th 2011, 06:32 AM
wik_chick88
Re: fourier transform of f(x)=e^(-ax^2) sin(bx)
im a little rusty on integration by parts, i know its \int u dv = uv - \int v du but i cant figure out what you have specified as u and dv above. also im still unsure as to how this helps me??
• Oct 9th 2011, 06:39 AM
girdav
Re: fourier transform of f(x)=e^(-ax^2) sin(bx)
I took $u(x)=xe^{-ax^2}$ and $v'(x)=e^{itx}$. Now you can solve the previous differential equation, since you can compute $G(0)$.