# Thread: fourier transform of f(x)=x, for 0<x<2

1. ## fourier transform of f(x)=x, for 0<x<2

Hi,

I am trying to find the Fourier transform of this function
$\displaystyle f(x) = \left\{ \begin{array}{cl} x & o \leq x \leq 2\\ 0 & \text{otherwise} \end{array} \right.$

What I have got so far:
$\displaystyle \hat{f}(u) = \frac{1}{\sqrt{2\pi}} \int_{0}^{2} x \cdot e^{-iux} dx$
$\displaystyle \sqrt{2\pi} \hat{f}(u)= -\int_{0}^{2}\frac{-1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}$
$\displaystyle =\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}$
$\displaystyle = \lbrack \frac{1}{u} e^{-iux} \rbrack_{0}^{2}+ \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}$
$\displaystyle = \left( \frac{1}{u}e^{-i2u} - \frac{1}{u} \right) + \left( \frac{-2}{iu} e^{-i2u} -0 \right)$

any thoughts would be very helpfull.

Thanks Alex.

2. Originally Posted by jeneverboy
Hi,

I am trying to find the Fourier transform of this function
$\displaystyle f(x) = \left\{ \begin{array}{cl} x & o \leq x \leq 2\\ 0 & \text{otherwise} \end{array} \right.$

What I have got so far:
$\displaystyle \hat{f}(u) = \frac{1}{\sqrt{2\pi}} \int_{0}^{2} x \cdot e^{-iux} dx$

$\displaystyle \sqrt{2\pi} \hat{f}(u)= -\int_{0}^{2}\frac{-1}{iu} e^{-iux} dx + \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}$

$\displaystyle =\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}$
Its OK up to here, but:

$\displaystyle \int_{0}^{2}\frac{1}{iu} e^{-iux} dx=\left[ \frac{1}{u^2}e^{-iux} \right]_0^2$

CB

3. allright, so
$\displaystyle \sqrt{2 \pi}\hat{f}(u)= \left[ \frac{1}{u^{2}} e^{-iux} \right]_{0}^{2}+ \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}$
$\displaystyle \sqrt{2 \pi}\hat{f}(u)= \left( \frac{1}{u^{2}}e^{-i2u} - \frac{1}{u^{2}} \right) + \left( \frac{-2}{iu} e^{-i2u} -0 \right)$
now Euler's formula can be used
$\displaystyle \boxed{ e^{ix}=cos x + i \; sinx }$
here
$\displaystyle e^{-2ui}=cos (-2u) + i \; sin(-2u)$
plug in equation
$\displaystyle \sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{1}{u^{2}} \left( cos(-2u)+i \; sin(-2u) \right) + \frac{-2}{iu} \left( cos(-2u) + i \; sin(-2u) \right)$
now I use $\displaystyle \frac{-2}{iu}=\frac{2i}{u}$ so equation becomes;
$\displaystyle \sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{1}{u^{2}} \left( cos(-2u)+i \; sin(-2u) \right) + \frac{2i}{u} \left( cos(-2u) + i \; sin(-2u) \right)$
rewrite
$\displaystyle \sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{cos(-2u)}{u^{2}} -\frac{2sin(-2u)}{u} + \frac{2i cos(-2u)}{u} + \frac{i sin(-2u)}{u^{2}}$
so now there is a real part and a imaginary part:
$\displaystyle \sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{cos(-2u)}{u^{2}} -\frac{2sin(-2u)}{u} + i \; \left( \frac{2 cos(-2u)}{u} + \frac{ sin(-2u)}{u^{2}} \right)$

is this correct so far?
does anyone have a suggestion how to simplify this?

thanks, Alex.

4. Originally Posted by jeneverboy
allright, so
$\displaystyle \sqrt{2 \pi}\hat{f}(u)= \left[ \frac{1}{u^{2}} e^{-iux} \right]_{0}^{2}+ \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}$

.
Just put:

$\displaystyle (\mathfrak{F}f)(u)=\frac{1}{\sqrt{2 \pi}}$$\displaystyle \left[ \frac{1}{u^2}(e^{-2ui}-1)+\frac{2i}{u}e^{-2ui} \right]$

and leave it there.

CB

5. Hi,

I need an expression with sin and cos, because now I must evaluate

$\displaystyle \int_{-\infty}^{\infty} \frac{sin(u)}{u} du$
and
$\displaystyle \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$

I can use the inverse transform here at a certain point $\displaystyle x_{0}$ which is a continuity point of $\displaystyle f$.

inversion formula:
$\displaystyle \lim_{a\rightarrow \infty} \frac{1}{\sqrt{2 \pi}} \int_{-a}^{a} \hat{f}(u)e^{iux_{0}} du = \frac{1}{2} (f(x_{0}^{+})+f(x_{0}^{-})) =f(x_{0})$
so this is $\displaystyle f(x_{0})$ because for instance at x=0 f is continuous, but for x=2 f is not continuous so there I can't use it.

But first I need an expression with cos and sin I think.

Alex.

6. Originally Posted by jeneverboy
Hi,

I need an expression with sin and cos, because now I must evaluate

$\displaystyle \int_{-\infty}^{\infty} \frac{sin(u)}{u} du$
and
$\displaystyle \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$

I can use the inverse transform here at a certain point $\displaystyle x_{0}$ which is a continuity point of $\displaystyle f$.

inversion formula:
$\displaystyle \lim_{a\rightarrow \infty} \frac{1}{\sqrt{2 \pi}} \int_{-a}^{a} \hat{f}(u)e^{iux_{0}} du = \frac{1}{2} (f(x_{0}^{+})+f(x_{0}^{-})) =f(x_{0})$
so this is $\displaystyle f(x_{0})$ because for instance at x=0 f is continuous, but for x=2 f is not continuous so there I can't use it.

But first I need an expression with cos and sin I think.

Alex.
Sorry but none of that make the least bit of sense.

Why don't you just post the quaestion you are really trying to address.

CB

7. ok here is the question I got

1a) Calculate the Fourier transform of the function
$\displaystyle f(x) = \left\{ \begin{array}{cl} x & o \leq x \leq 2\\ 0 & \text{otherwise} \end{array} \right.$

b)Evaluate
$\displaystyle \int_{-\infty}^{\infty} \frac{sin(u)}{u} du$
and
$\displaystyle \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$

8. Originally Posted by jeneverboy
ok here is the question I got

1a) Calculate the Fourier transform of the function
$\displaystyle f(x) = \left\{ \begin{array}{cl} x & o \leq x \leq 2\\ 0 & \text{otherwise} \end{array} \right.$

b)Evaluate
$\displaystyle \int_{-\infty}^{\infty} \frac{sin(u)}{u} du$
and
$\displaystyle \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$
For the second consider:

$\displaystyle \mathfrak{F}[f(x)](u) - \mathfrak{F}[f(-x)](u)$

For the first consider the use of the translation theorem.

CB