# Why is $\int_{-1}^1 (1-|t|) \cdot e^{-i \omega t} dt$ equal to $2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt$

I have a task where I should calculate the fourier transform of

$$\Delta(t) = \begin{cases} (1-|t|)& |t| \le 1 \\ 0 & |t| > 1 \end{cases}$$ The solution says, that $$\int_{-1}^1 (1-|t|) \cdot e^{-i\omega t} dt = 2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt$$

I understand that the factor must be $$2$$ and that we only need to integrate from 0 to 1 because $$\Delta(t)$$ is an even function. But i don't get the last transformation.

$$\int_{-1}^1 (1-|t|) \cdot e^{-i\omega t} dt = 2 \cdot \int_0^1 (1-t) \cdot e^{-i\omega t}\,dt = \mathop{???} = 2 \cdot \int_0^1 (1-t) \cdot \cos{\omega t}\, dt$$

Does anybody know how the transformation from $$e^{-i\omega t}$$ to $$\cos{\omega t}$$ works?

I would say that this are different functions, because normally i would say $$e^{-i\omega t} = \cos{\omega t} - i \sin{\omega t}$$

They have sort of buried the idea; $$\cos(\omega t)$$ is an even function of $$t$$ and $$\sin(\omega t)$$ is an odd function of $$t$$. If you use Euler's formula first before folding the integral in half it makes more sense.

To be more specific:

\begin{align*} \int_{-1}^{1} (1-|t|)e^{-i\omega t}dt &= \int_{-1}^{1} (1-|t|)(\cos(\omega t)-i\sin(\omega t))dt \\ &= \int_{-1}^{1} (1-|t|)\cos(\omega t)dt - i\int_{-1}^{1} (1-|t|)\sin(\omega t)dt \\ &= \int_{-1}^{1} (1-|t|)\cos(\omega t)dt +0 \end{align*}

• I tried to use eulers formula but i didn't get the desired result. – Sebi2020 Dec 7 '18 at 20:10
• @Sebi2020 The point is that the odd portion integrates out to $0$ over a symmetric interval around zero. That only leaves you with the even portion, and that folds in half to introduce the factor of $2$. – RandomMathGuy Dec 7 '18 at 20:11
• @Sebi2020 I have made an edit to make this more clear. – RandomMathGuy Dec 7 '18 at 20:19
• Ah, good point. I don't tried to partially calculate the cos and sin portions. – Sebi2020 Dec 7 '18 at 20:27

You wrote

$$\int_{-1}^1 (1-|t|) e^{-i\omega t} dt = 2\int_{0}^1 (1-t) e^{-i\omega t}dt$$

This is false, as $$e^{-i\omega t}$$ is not an even function. But all functions can be divided into an even part and an odd part. For $$e^{-i\omega t}$$, this is given by $$e^{-i\omega t} = \cos(\omega t) -i\sin(\omega t)$$. So we have $$\int_{-1}^1 (1-|t|) e^{-i\omega t} dt = \int_{-1}^1 (1-|t|) \cos(\omega t) dt - i\int_{-1}^1 (1-|t|) \sin(\omega t) dt \\ = 2\int_{-1}^1 (1-t) \cos(\omega t) dt,$$ where the sine integral is zero because its integrand is odd.

• Oh you're right, I only thought about (1-|t|) to being even and forgot, that this is not the case for $(1-|t|) \cdot e^{-i\omega t}$. – Sebi2020 Dec 7 '18 at 20:29