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We have to find the fourier complex transform of the following function $$f(x) = \begin{cases} 1-x^2 & |x| \leq 1 \\ 0 & \text{otherwise} \end{cases}$$

Here is my working.

(3) is the defination of Fourier transform which is defined as

$$\hat{f}(\omega)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-i \omega x} dx$$

Applying (3) and since $f(x)=0$ when $|x|>1$, then the value of integral when $x$ goes from $-\infty \rightarrow -1$ and $1 \rightarrow \infty$ will be 0 as well. Hence $$\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-1}^{1} (1-x^2) e^{-i\omega x} dx$$ Integrating by parts $$\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \left(\left[(1-x^2) \frac{e^{-i\omega x}}{-i\omega}\right]_{-1}^1 - \int_{-1}^{1}\frac{e^{-i\omega x}}{-i\omega} (-2x) dx\right)$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left(\left[(1-x^2) e^{-i\omega x}\right]_{-1}^1 + 2\int_{-1}^{1} e^{-i\omega x} (x) dx\right)$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left(\left[(1-x^2) e^{-i\omega x}\right]_{-1}^1 + 2\left[(x) \frac{e^{-i\omega x}}{-i\omega} \right]_{-1}^1 - 2\int_{-1}^{1} \frac{e^{-i\omega x}}{-i\omega} dx\right)$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left(\left[(1-x^2) e^{-i\omega x}\right]_{-1}^1 + 2\left[(x) \frac{e^{-i\omega x}}{-i\omega} \right]_{-1}^1 - 2\left[\frac{e^{-i\omega x}}{(-i\omega)^2} \right]_{-1}^1\right)$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left[e^{-i\omega x} \left(1-x^2 + \frac{2x}{-i\omega}- \frac{2}{(-i\omega)^2}\right)\right]_{-1}^1$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left[e^{-i\omega x} \left(1-x^2 + \frac{2(i\omega x+1)}{\omega^2}\right) \right]_{-1}^1$$ Applying limits, $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left[\left(e^{-i\omega} \left(0 + \frac{2(i\omega + 1)}{\omega^2}\right)\right) - \left(e^{i\omega} \left(0 + \frac{2(-i\omega +1)}{\omega^2}\right)\right)\right]$$ $$\hat{f}(\omega) = \frac{-1}{\sqrt{2\pi} i\omega} \left[e^{-i\omega} \left(\frac{2(i\omega + 1)}{\omega^2}\right) - e^{i\omega} \left(\frac{2(-i\omega +1)}{\omega^2}\right)\right]$$ $$\hat{f}(\omega) = \frac{-2}{\sqrt{2\pi} i\omega^3} \left(e^{-i\omega}(i\omega+1) - e^{i\omega}(1-i\omega)\right)$$ $$\hat{f}(\omega) = \frac{-2}{\sqrt{2\pi} i\omega^3} \left(i\omega(e^{-i\omega} - e^{i\omega}) + (e^{-i\omega} - e^{i\omega})\right)$$ Distributing $-i$ and multiplying/dividing with 2 to apply Euler's formula $$\hat{f}(\omega) = \frac{4}{\sqrt{2\pi} \omega^3} \left(i\omega(sin(\omega)) + (sin(\omega))\right)$$ $$\hat{f}(\omega) = \frac{2\sqrt{2}}{\sqrt{\pi} \omega^3} sin(\omega) \left(i\omega + 1\right)$$

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    $\begingroup$ Why do you think this is incorrect? $\endgroup$
    – knight5478
    Commented May 20 at 17:49
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    $\begingroup$ Something’s amiss, $\hat{f}$ ought to be real. $\endgroup$ Commented May 20 at 18:08

2 Answers 2

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You were right to detect that (since $f$ is even) $\hat f$ should be real. Your mistake occured very close to the end: $$\begin{align}\hat{f}(\omega)& = \frac{-2}{\sqrt{2\pi} i\omega^3} \left(e^{-i\omega}(i\omega+1) - e^{i\omega}(1-i\omega)\right)\\&=\frac{-2}{\sqrt{2\pi} i\omega^3} \left(i\omega(e^{-i\omega}\color{red}+e^{i\omega}) + (e^{-i\omega} - e^{i\omega})\right)\\ &=\frac4{\sqrt{2\pi}\omega^3} \left(-\omega\cos\omega+\sin\omega\right)\\ \end{align}$$

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Let $A=iw\int_{-1}^1x^2e^{ixw}dx$ and $B=iw\int_{-1}^1xe^{ixw}dx.$ Then $$A=[x^2e^{iwx}]_{-1}^1-2\int_{-1}^1xe^{ixw}dx=2i \sin w-2\frac{B}{iw}$$$$B=[xe^{iwx}]_{-1}^1-\int_{-1}^1e^{ixw}dx=2\cos w-2\frac{\sin w}{w}$$ $$\frac{A}{iw}=2\frac{\sin w}{w}+2\frac{B}{w^2}$$ $$\int_{-1}^1(1-x^2)e^{ixw}dx=2\frac{\sin w}{w}-\frac{A}{iw}=-2\frac{B}{w^2}$$$$=-4\frac{\cos w}{w^2}+4\frac{\sin w}{w^3}$$

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