Inverse Fourier transform with a $delta$-function integrand












2












$begingroup$


I'm self-studying math and trying to find the inverse Fourier transform of $frac{4+w^2}{1+w^2}(4pi * (delta(w-2)+delta(w+2)))$



Based on wolframalpha, the result is $32/5sqrt{2pi}cos(2t)$.



But I can't even find the fourier transform of $frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.






functions fourier-transform







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$endgroup$












  • $begingroup$
    Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
    $endgroup$
    – paul garrett
    Dec 7 '18 at 0:49
















2












$begingroup$


I'm self-studying math and trying to find the inverse Fourier transform of $frac{4+w^2}{1+w^2}(4pi * (delta(w-2)+delta(w+2)))$



Based on wolframalpha, the result is $32/5sqrt{2pi}cos(2t)$.



But I can't even find the fourier transform of $frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.






functions fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
    $endgroup$
    – paul garrett
    Dec 7 '18 at 0:49














2












2








2





$begingroup$


I'm self-studying math and trying to find the inverse Fourier transform of $frac{4+w^2}{1+w^2}(4pi * (delta(w-2)+delta(w+2)))$



Based on wolframalpha, the result is $32/5sqrt{2pi}cos(2t)$.



But I can't even find the fourier transform of $frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.






functions fourier-transform







share|cite|improve this question











$endgroup$




I'm self-studying math and trying to find the inverse Fourier transform of $frac{4+w^2}{1+w^2}(4pi * (delta(w-2)+delta(w+2)))$



Based on wolframalpha, the result is $32/5sqrt{2pi}cos(2t)$.



But I can't even find the fourier transform of $frac{4+w^2}{1+w^2}$ because there doesn't seem to have a Fourier transform pair in my table for $w^2$ in the numerator.






functions fourier-transform




functions fourier-transform






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share|cite|improve this question













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edited Dec 7 '18 at 1:00







drerD

















asked Dec 7 '18 at 0:44









drerDdrerD

1559




1559












  • $begingroup$
    Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
    $endgroup$
    – paul garrett
    Dec 7 '18 at 0:49


















  • $begingroup$
    Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
    $endgroup$
    – paul garrett
    Dec 7 '18 at 0:49
















$begingroup$
Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
$endgroup$
– paul garrett
Dec 7 '18 at 0:49




$begingroup$
Hint: subtracting the identically $1$ function from your $(4+w^2)/(1+w^2)$, you're left with $4/(1+w^2)$. The (tempered distribution) Fourier transform of $1$ is $delta$. The leftover ($4$ times) $1/(1+w^2)$ is in $L^1(mathbb R)$, and can easily be computed by complex analysis methods ("residues"). Up to some constants, it's $e^{-|x|}$.
$endgroup$
– paul garrett
Dec 7 '18 at 0:49










1 Answer
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$begingroup$

You don't need to know, just apply the definition



begin{eqnarray}
f(t) &=& frac{1}{2pi}int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}[4pidelta(omega - 2) + 4pidelta(omega + 2)] e^{iomega t}{rm d}omega \
&=& 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}4pidelta(omega - 2) e^{iomega t}{rm d}omega + 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2} delta(omega + 2) e^{iomega t}{rm d}omega \
&=& 2 frac{4 + 2^2}{1 + 2^2}e^{2it} + 2frac{4+2^2}{1 + 2^2}e^{-2it} \
&=&frac{32}{5}left(frac{e^{2it} + e^{-2it}}{2}right) = frac{32}{5}cos(2t)
end{eqnarray}






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    1 Answer
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    1 Answer
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    2












    $begingroup$

    You don't need to know, just apply the definition



    begin{eqnarray}
    f(t) &=& frac{1}{2pi}int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}[4pidelta(omega - 2) + 4pidelta(omega + 2)] e^{iomega t}{rm d}omega \
    &=& 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}4pidelta(omega - 2) e^{iomega t}{rm d}omega + 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2} delta(omega + 2) e^{iomega t}{rm d}omega \
    &=& 2 frac{4 + 2^2}{1 + 2^2}e^{2it} + 2frac{4+2^2}{1 + 2^2}e^{-2it} \
    &=&frac{32}{5}left(frac{e^{2it} + e^{-2it}}{2}right) = frac{32}{5}cos(2t)
    end{eqnarray}






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You don't need to know, just apply the definition



      begin{eqnarray}
      f(t) &=& frac{1}{2pi}int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}[4pidelta(omega - 2) + 4pidelta(omega + 2)] e^{iomega t}{rm d}omega \
      &=& 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}4pidelta(omega - 2) e^{iomega t}{rm d}omega + 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2} delta(omega + 2) e^{iomega t}{rm d}omega \
      &=& 2 frac{4 + 2^2}{1 + 2^2}e^{2it} + 2frac{4+2^2}{1 + 2^2}e^{-2it} \
      &=&frac{32}{5}left(frac{e^{2it} + e^{-2it}}{2}right) = frac{32}{5}cos(2t)
      end{eqnarray}






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You don't need to know, just apply the definition



        begin{eqnarray}
        f(t) &=& frac{1}{2pi}int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}[4pidelta(omega - 2) + 4pidelta(omega + 2)] e^{iomega t}{rm d}omega \
        &=& 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}4pidelta(omega - 2) e^{iomega t}{rm d}omega + 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2} delta(omega + 2) e^{iomega t}{rm d}omega \
        &=& 2 frac{4 + 2^2}{1 + 2^2}e^{2it} + 2frac{4+2^2}{1 + 2^2}e^{-2it} \
        &=&frac{32}{5}left(frac{e^{2it} + e^{-2it}}{2}right) = frac{32}{5}cos(2t)
        end{eqnarray}






        share|cite|improve this answer









        $endgroup$



        You don't need to know, just apply the definition



        begin{eqnarray}
        f(t) &=& frac{1}{2pi}int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}[4pidelta(omega - 2) + 4pidelta(omega + 2)] e^{iomega t}{rm d}omega \
        &=& 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2}4pidelta(omega - 2) e^{iomega t}{rm d}omega + 2int_{-infty}^{+infty} frac{4 + omega^2}{1 + omega^2} delta(omega + 2) e^{iomega t}{rm d}omega \
        &=& 2 frac{4 + 2^2}{1 + 2^2}e^{2it} + 2frac{4+2^2}{1 + 2^2}e^{-2it} \
        &=&frac{32}{5}left(frac{e^{2it} + e^{-2it}}{2}right) = frac{32}{5}cos(2t)
        end{eqnarray}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 0:55









        caveraccaverac

        14.6k31130




        14.6k31130






























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