Solve Intergrals Using Inverse Fourier Transform

Question

a)Find f(x), the insverse fourier transform of F(ω)

b) Does the fourier transform of f(x) equal to F(ω)?

c)use your answers to calculate these Integrals:

if I'm not mistaken the answer to a is:

as to b,c i not sure. please help!

Answer 1

$a)$ We have $$2\pi f\left(x\right) =\int_{\mathbb{R}}F\left(\omega\right)e^{i\omega x}\mathrm{d}\omega =\intop_{\omega=-L}^{L}\cos\left(a\omega\right)e^{i\omega x}\mathrm{d}\omega =\frac{1}{2}\left(\intop_{\omega=-L}^{L}e^{i\left(x+a\right)\omega}\mathrm{d}\omega+\intop_{\omega=-L}^{L}e^{i\left(x-a\right)\omega}\mathrm{d}\omega\right)$$ $$=\frac{1}{2}\left(\frac{e^{i\left(x+a\right)L}-e^{-i\left(x+a\right)L}}{i\left(x+a\right)}+\frac{e^{i\left(x-a\right)L}-e^{-i\left(x-a\right)L}}{i\left(x-a\right)}\right) =\frac{\sin\left(\left(x+a\right)L\right)}{x+a}+\frac{\sin\left(\left(x-a\right)L\right)}{x-a}$$ $$=\frac{\left(x-a\right)\sin\left(\left(x+a\right)L\right)+\left(x+a\right)\sin\left(\left(x-a\right)L\right)}{x^2-a^2}$$ $$=\frac{x\left(\sin\left(\left(x+a\right)L\right)+\sin\left(\left(x-a\right)L\right)\right)+a\left(\sin\left(\left(x-a\right)L\right)-\sin\left(\left(x+a\right)L\right)\right)}{x^2-a^2}$$ $$=\frac{2x\sin\left(Lx\right)\cos\left(aL\right)}{x^2-a^2} +\frac{2a\sin\left(aL\right)\cos\left(Lx\right)}{x^2-a^2}$$ where we used for the last equality the trigonometric relations $$\sin\left(a\right)-\sin\left(b\right) =2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right),$$ $$\sin\left(a\right)+\sin\left(b\right) =2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right).$$

$b)$ We have $$\int_{\mathbb{R}}|F\left(\omega\right)|^2\mathrm{d}\omega =\intop_{\omega=-L}^{L}\cos^2\left(a\omega\right)\mathrm{d}\omega =\intop_{\omega=-L}^{L}\frac{1+\cos\left(2a\omega\right)}{2}\mathrm{d}\omega =L+\frac{\sin\left(2aL\right)}{2a}<+\infty$$ hence $F\in\mathcal{L}^2\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right),\lambda\right)$, where $\mathcal{B}\left(\mathbb{R}\right)$ is the Borel's $\sigma$-algebra on $\mathbb{R}$ and $\lambda$ is the Lebesgue's measure on $\mathbb{R}$. Since the Fourier tansformation is one-to-one on $\mathcal{L}^2\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right),\lambda\right)$, we deduce that the Fourier transform of $f$ is $F$.

$c)$ For $L=\pi/a$, we have $$f\left(x\right)=-\frac{x\sin\left(\frac{\pi x}{a}\right)}{\pi\left(x^2-a^2\right)}$$ whence $$\int_{\mathbb{R}}\frac{x\sin\left(\frac{\pi x}{a}\right)\cos\left(\omega x\right)}{x^2-a^2}\mathrm{d}x =\Re\int_{\mathbb{R}}\frac{x\sin\left(\frac{\pi x}{a}\right)}{x^2-a^2}e^{-i\omega x}\mathrm{d}x =-\pi\Re\int_{\mathbb{R}}f\left(x\right)e^{-i\omega x}\mathrm{d}x$$ $$=-\pi\Re F\left(\omega\right)=-\pi\cos\left(a\omega\right)\mathfrak{1}_{_{\left[-\frac{\pi}{a},\frac{\pi}{a}\right]}}\left(\omega\right)$$ where $\mathfrak{1}_{_{\left[-\frac{\pi}{a},\frac{\pi}{a}\right]}}$ is the constant function equals to $1$ on $\left[-\frac{\pi}{a},\frac{\pi}{a}\right]$ and zero elsewhere.

For the second integral, do the same with $L=\pi/\left(2a\right)$.