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If the Fourier transform is defined by $\hat f( \xi)=\int_{-\infty}^{\infty}e^{-ix \xi}f(x)dx$.

How to calculate the Fourier transform of $$\begin{equation*} f(x)= \begin{cases} \frac{e^{ibx}}{\sqrt a} & \text{if $|x|\le a$,} \\ 0 &\text{if $|x|>a$.} \end{cases} \end{equation*}$$ The numbers $a$ and $b$ are positive.

I think when $|x|>a$, the Fourier transform is $0$. But when $|x|\le a$, I tried to calculate $\frac{1}{\sqrt a} \int_{-\infty}^{\infty}e^{ix(b-\xi)}dx$. I didn't learn complex analysis before, so I don't know how to calculate this integral. Can someone help me? Thanks so much!

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    $\begingroup$ You should be integrating from $-a$ to $a$, not $-\infty$ to $\infty$. $\endgroup$ – Bungo Dec 13 '14 at 17:47
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Note that $f(x)=0$ for all $|x|>a$, so

$$\hat{f}(\xi) = \int_{-\infty}^{\infty} e^{-\imath \, x \xi} f(x) \, = \int_{-a}^a e^{-\imath \, x \xi} f(x) \, dx.$$

Using the definition of $f$, we get

$$\hat{f}(\xi) = \frac{1}{\sqrt{a}} \int_{-a}^a e^{\imath \, b x} e^{-\imath \, x \xi} \, dx = \frac{1}{\sqrt{a}} \int_{-a}^a e^{\imath \, x(b-\xi)} \, dx.$$

Now use either

  • ... that $$\int_c^d e^{\imath \, x \alpha} \, dx = \frac{1}{\imath \alpha} \left( e^{\imath \, d \alpha}- e^{\imath \, c \alpha} \right)$$ for any $c \leq d$ and $\alpha \neq 0$
  • ... or write $$\int_{-a}^a e^{\imath \, x (b-\xi)} \, dx = \int_{-a}^a \cos(x (b-\xi)) \, dx + \imath \int_{-a}^a \sin(x (b-\xi)) \, dx$$ and integrate the terms separately.
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If the $i$ term of integral is throwing you for a loop, let $c = i(b-\xi)$. Then can you integrate $$ \frac{1}{\sqrt{a}}\int_{-\infty}^{\infty}e^{cx}dx\mbox{?} $$ Of course once you do integrate, you will need to remember to back substitute for $c$.

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  • $\begingroup$ Hi, dustin. There is something I feel confused. When can the $i$ can be viewed as a constant? Why in the case $\int_{-\infty}^{\infty}e^{-(x+i\xi)^2}dx$, $i$ cannot be viewed as a constant? Thanks! $\endgroup$ – Sherry Dec 13 '14 at 17:46
  • $\begingroup$ @Sherry $i$ is a constant as long as it is $i = \sqrt{-1}$. In the case you mention, you have $i$ mixed in $x$ so it doesn't just pull out. $\endgroup$ – dustin Dec 13 '14 at 17:48

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