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$$f(x)=\frac{e^{2ix}}{1+x^2}+3p_2(x-1)$$


$$p_2(x)=\begin{cases} 1 \qquad x \in [-1,1] \\ 0 \qquad x \notin [-1,1] \end{cases}$$


$$\int_{-\infty}^{+\infty} p_2(x) \ e^{-i\omega x} \ dx=2 \ \frac{\sin(\omega )}{\omega} $$
So: $$\int_{-\infty}^{+\infty} 3 p_2(x-1) \ e^{-i\omega x} \ dx=6 \ \frac{\sin(\omega )}{\omega} \ e^{-i \omega }$$


$$\int_{-\infty}^{+\infty} \frac{e^{2ix}}{1+x^2} \ e^{-i\omega x} \ dx$$

Using Residue theorem:

$$\int_{-\infty}^{+\infty} \frac{e^{2ix}}{1+x^2} \ e^{-i\omega x} \ dx=2\pi i \ \ \lim_{x\rightarrow i} \ \frac{e^{-i\omega x+2ix}}{x+i}=2 \pi i \ \ \frac{e^\omega \ e^{-2}}{2i}=\pi \ e^\omega \ e^{-2} $$


$$F(\omega)=\int_{-\infty}^{+\infty} f(x) \ e^{-i\omega x} \ dx=\pi \ e^\omega \ e^{-2}+6 \ \frac{\sin(\omega )}{\omega} \ e^{-i \omega }$$


Is it correct?

Which are the conditions I have to verify to establish if $F$ inverse Fourier transform is possible?
Should I verify $F \in L^1(\mathbb{R})$?

Thanks!

  • 1
    if $\int_{\mathbb{R}}\tilde{f}(\omega)d\omega$ exists the inverse FT is well defined. So yes, you are right – 2017-01-11
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    since $||f(x)||_1\geq ||f(x)||_p$ any other norm also does the job – 2017-01-11

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