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

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

i need to find the fourier transform of $\displaystyle f(x)=e^{-ax^2} sin(bx)$

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

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

im stuck as to where to go from here! judging by our notes the next step should be
$\displaystyle = \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \sqrt{\frac{\pi}{a}} \int^{\infty}_{-\infty} e^{-\frac{(b+\lambda)^2}{4a}} - e^{-\frac{(b-\lambda)^2}{4a}}$
but i have no idea why this should be (and not even sure its the right step!)

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

You will be able to conclude if you know the value of $\displaystyle I(a,b):=\int_{-\infty}^{+\infty}e^{-ax^2}e^{ibx}dx$. If you fix $\displaystyle a$, differentiate this with respect to $\displaystyle b$, and integrate by parts, you will find a differential equation.

3. ## 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?

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

Fix $\displaystyle a$ and put $\displaystyle G(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}dx$. We have $\displaystyle G'(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}ixdx=i\left[-e^{-ax^2}\frac 2ae^{itx}\right]_{-\infty}^{+\infty}+\frac{2i}a\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}itdx$ hence $\displaystyle G'(t)=-\frac{2t}aG(t)$.

5. ## 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??

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

I took $\displaystyle u(x)=xe^{-ax^2}$ and $\displaystyle v'(x)=e^{itx}$. Now you can solve the previous differential equation, since you can compute $\displaystyle G(0)$.

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# find fourier transform of e(-ax2)

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