Question

This was on the test I just took.

Find the fourier transform of \[f(x)=\left[\begin{matrix}|x| & |x|<1 \\ 0 & |x| \ge 1\end{matrix}\right]\]
\[{\hat{f}(x)}=\int\limits_{- \infty}^{\infty}f(x)e^{-i \xi x}dx\]

Answer 1

i think u just need to calculate this integral

\[{\hat{f}(x)}=\int\limits_{- 1}^{1}|x|e^{-i \xi x}dx\]

Answer 2

I hope so, because that's what I did :)

Answer 3

Did you get this answer

\[
\int_{-1}^0 -x e^{-i \mu \, x}dx + \int_0^1 x e^{-i \mu \, x}dx\\
\frac{-1+e^{i \mu } (1-i \mu
)}{\mu ^2}+\frac{-1+e^{-i \mu
} (1+i \mu )}{\mu ^2}=\\
\frac{e^{-i \mu } \left(-1+e^{i
\mu }\right) \left(-i e^{i
\mu } \mu -i \mu +e^{i \mu
}-1\right)}{\mu ^2}
\]
\

Answer 4

The first thing I did was \[\int\limits_{-1}^{1}|x|(\cos( \xi x)+isin(\xi x))dx\]
via euler's identity\[e^{ix}=\cos(x)-isin(x)\]
\[\int\limits_{-1}^{1}(|x|\cos( \xi x)+|x|isin(\xi x))dx=\int\limits_{-1}^{1}|x|\cos( \xi x)dx\]

Answer 5

and those plusses are supposed to be minuses