Changing the limits of an integral of a function

Question

I want to find the Fourier transform of the following function:

How would I integrate this function between $-\infty$ and $\infty$ multiplied by exp(-ikx)

$$ f(x) =\cases{ 1 &if 0 $\leq x \leq 1$\\ 0 &if $x>1$} $$

I thought of splitting the integral up into three separate integrals:

1. between $-\infty$ and $0$
2. between $0$ and $1$
3. between $1$ and $\infty$

However the function isn't defined between $-\infty$ and $0$ so I'm not sure this would work?

Also, is it possible to change the integral from $-\infty$ to $\infty$ into twice the integral going from $0$ to $1$ with use of the fact that the function is even, or the integral is zero if the function is odd?

Although, I cant seem to calculate whether the function is odd or even.

Could someone give me some guidance? Thanks!

Answer 1

If a function is not defined on some domain with non-zero size, you can't integrate that function over that domain. So an undefined point is fine, but an undefined region is not. Your problem is not well-formed.

Answer 2

So it doesn’t make any sense to integrate from $-\infty$ if $f$ is only defined for $x\ge 0$ so there are two things we can do: either set $f$ to be 0 for negative $x$ or integrate from 0. Otherwise we would need to reformulate the problem so it does make sense. They both have the same effect and so we split the integral into three (or two) parts. One is integarating 1 from 0 to 1 so has the value of 1. The other is the integral of 0 so has the value 0. This the whole integral is 1.

Draw a graph of $f$ and work out the area under the axis. You clearly have a square of area 1.

You suggested the following: $$\int_{-\infty}^\infty f = 2\int_0^\infty f, $$ But this only works if $f(x)=f(-x)$ for all $x$ (ie $f$ is even) and your function does not satisfy this.

Answer 3

However the function isn't defined between $-\infty$ and $0$ so I'm not sure this would work?

It won't work.

Also, is it possible to change the integral from $-\infty$ to $\infty$ into twice the integral going from $0$ to $1$ with use of the fact that the function is even, or the integral is zero if the function is odd?

Or anything else if the function is neither even nor odd, as is the case for most funcitons (under some definitions of "most").

Although, I cant seem to calculate whether the function is odd or even.

Which makes sense, considering evenness and oddness for a function relates the function values at positive input with the function values for negative input, and you haven't been told anything about the function values at negative inputs.