Fourier transform of a triangular pulse I've been practicing some Fourier transform questions and stumbled on the following one.



To start off, I defined the Fourier transform for this function by taking integral from $-\tau$ to $0$ and $0$ to $\tau$, as shown below

from that, I evaluated the first integral and got the following result

then followed by the second integral

From the two integrals, I tried to solve for $X(\omega)$ by summing the two integrals

Now, this is where I got stuck. I am not sure how to express this as the required answer. Am I missing some small details or is it some basic algebra?
 A: Sinc function is tricky, because there are two of them. It seems your book uses the convention
$$\operatorname{sinc} x = \frac{\sin (\pi x)}{\pi x}$$
The desired answer is 
$$X(\tau) = \tau\frac{\sin^2 (\omega \tau/2)}{(\omega \tau/2)^2}  = \frac{4}{\omega^2 \tau }\sin^2 (\omega \tau/2) =\frac{2}{\omega^2 \tau }(1-\cos \omega \tau) $$
Which is what you have, since $e^{j\omega\tau}+e^{-j\omega\tau}=2\cos  \omega\tau$.
A: Continue from what you have got:
$$\frac{2-(e^{j\omega \tau} + e^{-j\omega \tau)}}{\omega^2 \tau} = \frac{2-2\cos(\omega \tau)}{\omega^2 \tau} = \frac{2(1-\cos(\omega \tau))}{\omega^2 \tau} $$
Recall that
$$1-\cos(\omega \tau) = 2 \sin^2 (\frac{\omega \tau}{2})$$
Thus
$$\frac{2(1-\cos(\omega \tau))}{\omega^2 \tau} = \frac{4\sin^2 (\frac{\omega \tau}{2})}{\omega^2 \tau}$$
Rearrange by multiplying the numerator and the denominator by $\frac{\tau}{4}$
$$\frac{4\sin^2 (\frac{\omega \tau}{2})}{\omega^2 \tau} \cdot \frac{\frac{\tau}{4}}{\frac{\tau}{4}} = \tau \left[\frac{\sin (\frac{\omega \tau}{2})}{\frac{\omega \tau}{2}}\right]^2$$
And recall that
$$\mathrm{sinc}(\frac{\omega \tau}{2}) = \frac{\sin (\frac{\omega \tau}{2})}{\frac{\omega \tau}{2}}$$
Therefore, 
$$\mathfrak{F}[\mathrm{tri}(\frac{t}{\tau}] = \tau  \mathrm{sinc}^2(\frac{\omega \tau}{2})$$
A: A simpler way to arrive at the expression involving the cosine term is to consider the symmetry of the triangular pulse. Since it is an even function, multiplication by exp(-jwt) is equivalent to multiplying by coswt, since the sine term will go to zero. Then we get lesser number of integrals to evaluate and the same expression involving [1-cos(omega.tau)]can be obtained much more easily.
Hope this helps.
