Anas.P
  • Anas.P
fourier traorm question please help
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Anas.P
  • Anas.P
Find the fourier sine transform of the function\[f(x)=e^{|x|}\]
anonymous
  • anonymous
\[\large \begin{align*} \mathcal{F}\left\{e^{|x|}\right\}&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{|x|}e^{-i\xi x}\,dx\\[1ex] &=\frac{1}{\sqrt{2\pi}}\left(\int_0^\infty e^{x}e^{-i\xi x}\,dx+\int_{-\infty}^0 e^{-x}e^{-i\xi x}\,dx\right)\\[1ex] &=\frac{1}{\sqrt{2\pi}}\left(\int_0^\infty e^{x}e^{-i\xi x}\,dx+\int_0^\infty e^{x}e^{i\xi x}\,dx\right)\\[1ex] &=\frac{2}{\sqrt{2\pi}}\int_0^\infty e^{x}\frac{e^{i\xi x}+e^{-i\xi x}}{2}\,dx\\[1ex] &=\frac{2}{\sqrt{2\pi}}\int_0^\infty e^{x}\cos\xi x\,dx\\[1ex] \end{align*}\] Notice that parameterizing the above integral as \[\large I(s)=\int_0^\infty e^{-sx}\cos\xi x\,dx\] gives the Laplace transform of \(\cos \xi x\), where in this case \(s=-1\).
anonymous
  • anonymous
Whoops, you had asked for the sine transform... Sorry about that. In any case, you can proceed similarly, splitting the domain at \(x=0\).

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IrishBoy123
  • IrishBoy123
\[\mathcal{F_{s}}\left\{e^{|x|} \right\} =-\frac{1}{\sqrt{2\pi}} \mathcal {Im} \ \left[ \ \int_{-\infty}^\infty e^{|x|}e^{-i\xi x}\,dx\ \right] \] \[=-\frac{2}{\sqrt{2\pi}} \mathcal{Im} \ \left[ \int_0^\infty e^{x}\cos\xi x\,dx\ \right]\] \[=0\]
Anas.P
  • Anas.P
but the sine transform has limits 0 to infinity and therefore no -x should appear. Am i correct?
IrishBoy123
  • IrishBoy123
the sine transform has the limits as per the definition of the complete fourier integral, ie \(-\infty \lt t \lt \infty\) however, if the function is odd, you can make that leap, and make it an \(0 \lt t \lt \infty\) integration ..... times 2 just as, if the function is even, you can do the same thing in reverse. which is just one of the interesting conclusions from sith's integration moreover, an even function such as \(e^{|x|}\) will have a real transform and so the imaginary term will be zero. so the answer to this was clear from the start. if you go back to those original ideas of integral calculus, where you were summing little rectangles..... :p but now you first multiplying the areas of those little rectangles by either sine or cos, and one is odd and one is even
Anas.P
  • Anas.P
thanks....

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