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I attempted this problem, but my final answer does not seem simplified enough. The problem is that I don't know how to simplify it more and I lack the experience to be able to tell when it's ok to stop simplifying.

My attempt:

$f(x)=|x|, |x|<1 \quad ; \quad =0 \quad $everywhere else

$$g(a)=\frac{1}{2\pi}\int_{-1}^1 |x|e^{-iax}dx$$

$$=\frac{1}{\pi}\int_0^1 xe^{-iax}dx$$

$$=\frac{1}{\pi} \left [\frac{xe^{-iax}}{-ia}|_0^1 -\int_0^1 \frac{e^{-iax}}{-ia} \right ]$$

$$=-\frac{1}{i\pi a} \left [e^{-ia} + \frac{e^{-ia}-1}{ia} \right ]$$

I'm new to Fourier transforming, so I'm not sure what a useable final answer looks like. Can this be simplified further?

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  • $\begingroup$ You should have left it as an integral from $-1$ to $1$ and simple your answer into trig functions. $\endgroup$ Commented Oct 20, 2016 at 3:30
  • $\begingroup$ @JackyChong Why? This is an even function across the y-axis, so can't I just integrate from $0$ to $1$ and multiply by $2$? $\endgroup$ Commented Oct 20, 2016 at 3:31
  • $\begingroup$ Sure, but you could simplify your expression if you just stick to integration from $-1$ to $1$. $\endgroup$ Commented Oct 20, 2016 at 3:32
  • $\begingroup$ But then how do I integrate the absolute value of $x$? $\endgroup$ Commented Oct 20, 2016 at 3:33
  • $\begingroup$ Let me write a short note in the answer section $\endgroup$ Commented Oct 20, 2016 at 3:33

2 Answers 2

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Observe \begin{align} \frac{1}{2\pi}\int^1_{-1}|x|e^{-iax}\ dx =&\ \frac{1}{2\pi} \int^1_0 x e^{-iax} dx + \frac{1}{2\pi} \int^0_{-1}-xe^{-ia x}\ dx\\ =&\ \frac{1}{2\pi} \left[\frac{xe^{-iax}}{-ia}\bigg|^1_0+\int^1_0 \frac{e^{-iax}}{ia}\ dx \right]- \frac{1}{2\pi}\left[\frac{xe^{iax}}{-ia}\bigg|^0_{-1}+\int^0_{-1}\frac{e^{iax}}{ia}\ dx\right]\\ =&\ \frac{1}{2\pi}\left[\frac{-1+(1+i a) e^{-i a}}{a^2}\right]- \frac{1}{2\pi}\left[ \frac{1+i (i+a) e^{i a}}{a^2}\right]\\ =&\ \frac{-1}{\pi a^2}+\frac{1}{2\pi} \frac{(1+ia)e^{-ia}-(-1+ia)e^{ia}}{a^2}\\ =&\ \frac{-1}{\pi a^2} +\frac{1}{2\pi}\frac{e^{ia}+e^{-ia}}{a^2}+\frac{1}{\pi}\frac{e^{ia}-e^{-ia}}{2ia} = \frac{-1}{\pi a^2}+\frac{\cos a}{\pi a^2}+\frac{a\sin a}{\pi a^2} \end{align}

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You can simplify it further by using trig functions instead of the Euler notation. I would not recommend integrating from -1 to 1 since (you already seem to know) that integrating $|x|$ would have you dealing with the sgn(x) function and that's over-complicating the problem.

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