1
$\begingroup$

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

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

Edit:

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

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

where

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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,

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.