The function is, $f(x) = \begin{cases}0 & x < 0\\e^{-x} & x > 0\end{cases}$

I have to find the fourier integral representation and hence show that

$$\int_{0}^{\infty} \frac{\cos\omega{x}+\omega{\sin\omega{x}}}{1+\omega^{2}}dw = \begin{cases}0 & x < 0\\\frac{\pi}{2} & x = 0\\\pi{e^{-x}} & x>0\end{cases}$$


The fourier integral representation of a function is defined as follows:

$$ f(x)=\int_{0}^{\infty} [A(w)coswx+B(w)sinwx] dw $$


$$A(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)coswv]dv$$ $$B(w)= \frac{1}{\pi}\int_{-\infty}^{\infty} [f(v)sinwv]dv$$

  • $\begingroup$ By fourier integral do you mean fourier transform or something else? $\endgroup$ – TSF Feb 5 '18 at 12:31
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Feb 5 '18 at 12:33

Take the fourier transform using the definition as an integral,

$$\mathcal{F} (f(x)) = \int_{\mathbb{R}} f(x)e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x}e^{-2\pi i x\xi}dx=\int_0^\infty e^{-x(2\pi i\xi + 1)}dx$$

Now just evaluate the integral to get,

$$\mathcal{F} (f(x)) = \frac{e^{-x(2\pi i \xi +1)}}{-(2\pi i \xi+1)}\biggr|_0^\infty = \frac{1}{(2\pi i \xi +1)}$$

  • $\begingroup$ We haven't been taught about fourier transforms yet. Although the instructor did mention it. $\endgroup$ – Atkrista Feb 5 '18 at 13:26
  • $\begingroup$ In this case, using the definitions you give above, you can evaluate those integrals using integration by parts several times. $\endgroup$ – TSF Feb 5 '18 at 16:19

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