Fourier transform between variables of different domains

[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. Homework Statement
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 $[0,1]$.

Homework Equations

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

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.

 

[Ray Vickson]

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. Homework Statement
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 $[0,1]$.

Homework Equations

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

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.

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, $f(x) = f(x),\, x \neq 0,1$.