# Homework Help: Fourier transform between variables of different domains

1. Dec 19, 2017

### alec_grunn

I'm doing a research project over the summer, and need some help understanding how to construct an inverse Fourier transform (I have v. little prior experience with them).

1. The problem statement, all variables and given/known data

I know the explicit form of $q(x)$, where
$$q(x) = \frac{M}{2 \pi} \int _{- \infty}^{\infty} dz e^{-iMxz} C_q(z)$$
and want to find $C_q(z)$ using an inverse Fourier transform. As far as I can tell, there's no simple relationship between $z$ and $x$. And the domain of $x$ is $[0,1]$.

2. Relevant Equations

Explicit form of $q(x)$: $q(x) = x^{1/5}(1-x)^3$.

3. The attempt at a solution

I thought I would start with a substitution, since $z$ and $M$ are independent: $\mu = Mz$. Therefore,
$$q(x) = \frac{1}{2 \pi} \int _{- \infty}^{\infty} d\mu e^{-ix\mu} \tilde{C}_{q}(\mu)$$
And from this relation I use the inverse Fourier transform to get
$$\tilde{C}_{q}(\mu) = \int _{0}^{1} dx e^{ix\mu} q(x)$$
$$\Rightarrow \quad C_{q}(z) = \int _{0}^{1} dx e^{iMxz} q(x)$$
Is this reasoning sound? Any help is appreciated.

2. Dec 19, 2017

### Ray Vickson

Do you mean
$$q(x) = \begin{cases} 0 & x < 0 \\ x^{1/5}(1-x)^3 & 0 \leq x \leq 1 \\ 0 & x > 1 \end{cases}$$
If so, the Fourier transform is
$$F(k) = \int_0^1 e^{ikx} q(x) \, dx$$
and the inverse Fourier transform is
$$q(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-ikx} F(k) \, dk$$

If you do everything correctly the inverse Fourier transform should evaluate to $0$ if $x < 0$ or $x > 1$, and to $x^{1/5}(1-x)^3$ if $0 \leq x \leq 1$.

To see what is happening, look at the simpler example of
$$f(x) = \begin{cases} 0, & x< 0 \\ 1, & 0 \leq x \leq 1 \\ 0, & x > 1 \end{cases}$$
Computing $g(k) = \int_0^1 e^{i k x} \, dx$ is easy, and inverting to get
$$F(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g(k) e^{-ikx} \, dk$$
is also relatively straightforward. You will find that $F(x) = 0$ for $x<0 \; \text{or} \; x>1$, and $F(x) = 1$ for $0 < x < 1$.

However, the values of $F(x)$ at $x=0$ and at $x=1$ may differ from $f(0)$ and $f(1)$ for the usual reasons about the values of Fourier series/integrals at discontinuity points.

In fact:
$$F(x) = \begin{cases} 0,& x < 0 \\ 1/2, & x = 0 \\ 1, & 0 < x < 1 \\ 1/2, & x=1 \\ 0 , & x > 1 \end{cases}$$
Thus, $f(x) = f(x),\, x \neq 0,1$.

Last edited: Dec 19, 2017