Find the fourier transform of$$e^{-ax^2}$$
$$F[f(\xi)]=\int\limits_{-\infty}^{\infty}f(x)e^{-i \xi x}dx$$

You will need to know this fact:$$2 \int\limits_{0}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$

I can get to
$$2\int\limits_{0}^{\infty}e^{-ax^2}\cos(\xi x)dx$$

Question

Find the fourier transform of$$e^{-ax^2}$$
$$F[f(\xi)]=\int\limits_{-\infty}^{\infty}f(x)e^{-i \xi x}dx$$

You will need to know this fact:$$2 \int\limits_{0}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$

I can get to
$$2\int\limits_{0}^{\infty}e^{-ax^2}\cos(\xi x)dx$$

anonymous · Answer

u have to calculate last integral?

anonymous · Answer

Yup

anonymous · Answer

Ans is attached btw.

anonymous · Answer

The next step has to do with differentiating the inegral:

$$L(\xi)=2\int\limits_{0}^{\infty}e^{-ax^2}\cos(\xi x)dx$$$$L'(\xi)=-2\int\limits_{0}^{\infty}xe^{-ax^2}\sin(\xi x)dx$$

But not really sure how that got differentiated.

anonymous · Answer

$$F(a,b)=\int\limits_{0}^{\infty}e^{-ax^2}\cos(b x)dx$$
$$\frac{\partial F}{\partial b}=-\int\limits_{0}^{\infty}xe^{-ax^2}\sin(b x)dx$$
now use integration by parts to show that $$\frac{\partial F}{\partial b}=-\frac{b}{2a} F$$ 

solve for $F$

anonymous · Answer

I knew that hehe, brain fart

anonymous · Answer

lol...:)

anonymous · Answer

Made it through... such a strange problem, turned in a diffeq... gonna be a fun test :)