Finding inverse Fourier transform of $frac{1}{(1+iw)^2}$












0












$begingroup$



Find the inverse Fourier transform of the function $$ frac{1}{(1+iw)^2}
$$




So I know the inverse is given by the integral $$ frac{1}{sqrt{2 pi}} int_{-infty}^infty frac{e^{iwx}}{(1+iw)^2}dw $$ but I don't know how to go about solving this.






fourier-analysis improper-integrals fourier-transform







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$endgroup$












  • $begingroup$
    Do you know the residue theorem?
    $endgroup$
    – Fabian
    Dec 1 '18 at 17:30










  • $begingroup$
    Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
    $endgroup$
    – Tesseract
    Dec 1 '18 at 17:47










  • $begingroup$
    This might strongly depend on the different ways defining the Fourier transform.
    $endgroup$
    – Jonas Lenz
    Dec 1 '18 at 19:41










  • $begingroup$
    @JonasLenz Assume its given by the integral in my post.
    $endgroup$
    – Pame
    Dec 2 '18 at 9:44
















0












$begingroup$



Find the inverse Fourier transform of the function $$ frac{1}{(1+iw)^2}
$$




So I know the inverse is given by the integral $$ frac{1}{sqrt{2 pi}} int_{-infty}^infty frac{e^{iwx}}{(1+iw)^2}dw $$ but I don't know how to go about solving this.






fourier-analysis improper-integrals fourier-transform







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know the residue theorem?
    $endgroup$
    – Fabian
    Dec 1 '18 at 17:30










  • $begingroup$
    Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
    $endgroup$
    – Tesseract
    Dec 1 '18 at 17:47










  • $begingroup$
    This might strongly depend on the different ways defining the Fourier transform.
    $endgroup$
    – Jonas Lenz
    Dec 1 '18 at 19:41










  • $begingroup$
    @JonasLenz Assume its given by the integral in my post.
    $endgroup$
    – Pame
    Dec 2 '18 at 9:44














0












0








0





$begingroup$



Find the inverse Fourier transform of the function $$ frac{1}{(1+iw)^2}
$$




So I know the inverse is given by the integral $$ frac{1}{sqrt{2 pi}} int_{-infty}^infty frac{e^{iwx}}{(1+iw)^2}dw $$ but I don't know how to go about solving this.






fourier-analysis improper-integrals fourier-transform







share|cite|improve this question











$endgroup$





Find the inverse Fourier transform of the function $$ frac{1}{(1+iw)^2}
$$




So I know the inverse is given by the integral $$ frac{1}{sqrt{2 pi}} int_{-infty}^infty frac{e^{iwx}}{(1+iw)^2}dw $$ but I don't know how to go about solving this.






fourier-analysis improper-integrals fourier-transform




fourier-analysis improper-integrals fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 19:21







Pame

















asked Dec 1 '18 at 16:51









PamePame

35817




35817












  • $begingroup$
    Do you know the residue theorem?
    $endgroup$
    – Fabian
    Dec 1 '18 at 17:30










  • $begingroup$
    Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
    $endgroup$
    – Tesseract
    Dec 1 '18 at 17:47










  • $begingroup$
    This might strongly depend on the different ways defining the Fourier transform.
    $endgroup$
    – Jonas Lenz
    Dec 1 '18 at 19:41










  • $begingroup$
    @JonasLenz Assume its given by the integral in my post.
    $endgroup$
    – Pame
    Dec 2 '18 at 9:44


















  • $begingroup$
    Do you know the residue theorem?
    $endgroup$
    – Fabian
    Dec 1 '18 at 17:30










  • $begingroup$
    Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
    $endgroup$
    – Tesseract
    Dec 1 '18 at 17:47










  • $begingroup$
    This might strongly depend on the different ways defining the Fourier transform.
    $endgroup$
    – Jonas Lenz
    Dec 1 '18 at 19:41










  • $begingroup$
    @JonasLenz Assume its given by the integral in my post.
    $endgroup$
    – Pame
    Dec 2 '18 at 9:44
















$begingroup$
Do you know the residue theorem?
$endgroup$
– Fabian
Dec 1 '18 at 17:30




$begingroup$
Do you know the residue theorem?
$endgroup$
– Fabian
Dec 1 '18 at 17:30












$begingroup$
Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
$endgroup$
– Tesseract
Dec 1 '18 at 17:47




$begingroup$
Note that the $e^{-iwx}$ should be $e^{iwx}$ for an inverse transform.
$endgroup$
– Tesseract
Dec 1 '18 at 17:47












$begingroup$
This might strongly depend on the different ways defining the Fourier transform.
$endgroup$
– Jonas Lenz
Dec 1 '18 at 19:41




$begingroup$
This might strongly depend on the different ways defining the Fourier transform.
$endgroup$
– Jonas Lenz
Dec 1 '18 at 19:41












$begingroup$
@JonasLenz Assume its given by the integral in my post.
$endgroup$
– Pame
Dec 2 '18 at 9:44




$begingroup$
@JonasLenz Assume its given by the integral in my post.
$endgroup$
– Pame
Dec 2 '18 at 9:44










1 Answer
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0












$begingroup$

A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.



Observe that



$$dfrac{1}{(1+iomega)^2} = idfrac{d}{domega} dfrac{1}{1+iomega}$$



and note that the Fourier Transform has a derivative theorem



$$mathscr{F}left{x^n f(x)right}=mathscr{F}left{x^n mathscr{F}^{-1}left{ F(omega)right}right}= i^ndfrac{d^n}{domega^n} F(omega)$$



A table lookup of Fourier Transforms yields



$$mathscr{F}left{e^{-ax}H(x)right} = dfrac{1}{sqrt{2pi}}cdotdfrac{1}{a+iomega}$$



where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)



You can now say that



$$begin{align*}mathscr{F}^{-1}left{dfrac{1}{(1+iomega)^2}right} &= mathscr{F}^{-1}left{ idfrac{d}{domega} dfrac{1}{1+iomega}right}\
\
&= xmathscr{F}^{-1}left{ dfrac{1}{1+iomega}right}\
\
&= sqrt{2pi}xmathscr{F}^{-1}left{dfrac{1}{sqrt{2pi}}cdot dfrac{1}{1+iomega}right}\
\
&= sqrt{2pi}xe^{-x}H(x)\
end{align*}$$






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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    0












    $begingroup$

    A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.



    Observe that



    $$dfrac{1}{(1+iomega)^2} = idfrac{d}{domega} dfrac{1}{1+iomega}$$



    and note that the Fourier Transform has a derivative theorem



    $$mathscr{F}left{x^n f(x)right}=mathscr{F}left{x^n mathscr{F}^{-1}left{ F(omega)right}right}= i^ndfrac{d^n}{domega^n} F(omega)$$



    A table lookup of Fourier Transforms yields



    $$mathscr{F}left{e^{-ax}H(x)right} = dfrac{1}{sqrt{2pi}}cdotdfrac{1}{a+iomega}$$



    where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)



    You can now say that



    $$begin{align*}mathscr{F}^{-1}left{dfrac{1}{(1+iomega)^2}right} &= mathscr{F}^{-1}left{ idfrac{d}{domega} dfrac{1}{1+iomega}right}\
    \
    &= xmathscr{F}^{-1}left{ dfrac{1}{1+iomega}right}\
    \
    &= sqrt{2pi}xmathscr{F}^{-1}left{dfrac{1}{sqrt{2pi}}cdot dfrac{1}{1+iomega}right}\
    \
    &= sqrt{2pi}xe^{-x}H(x)\
    end{align*}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.



      Observe that



      $$dfrac{1}{(1+iomega)^2} = idfrac{d}{domega} dfrac{1}{1+iomega}$$



      and note that the Fourier Transform has a derivative theorem



      $$mathscr{F}left{x^n f(x)right}=mathscr{F}left{x^n mathscr{F}^{-1}left{ F(omega)right}right}= i^ndfrac{d^n}{domega^n} F(omega)$$



      A table lookup of Fourier Transforms yields



      $$mathscr{F}left{e^{-ax}H(x)right} = dfrac{1}{sqrt{2pi}}cdotdfrac{1}{a+iomega}$$



      where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)



      You can now say that



      $$begin{align*}mathscr{F}^{-1}left{dfrac{1}{(1+iomega)^2}right} &= mathscr{F}^{-1}left{ idfrac{d}{domega} dfrac{1}{1+iomega}right}\
      \
      &= xmathscr{F}^{-1}left{ dfrac{1}{1+iomega}right}\
      \
      &= sqrt{2pi}xmathscr{F}^{-1}left{dfrac{1}{sqrt{2pi}}cdot dfrac{1}{1+iomega}right}\
      \
      &= sqrt{2pi}xe^{-x}H(x)\
      end{align*}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.



        Observe that



        $$dfrac{1}{(1+iomega)^2} = idfrac{d}{domega} dfrac{1}{1+iomega}$$



        and note that the Fourier Transform has a derivative theorem



        $$mathscr{F}left{x^n f(x)right}=mathscr{F}left{x^n mathscr{F}^{-1}left{ F(omega)right}right}= i^ndfrac{d^n}{domega^n} F(omega)$$



        A table lookup of Fourier Transforms yields



        $$mathscr{F}left{e^{-ax}H(x)right} = dfrac{1}{sqrt{2pi}}cdotdfrac{1}{a+iomega}$$



        where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)



        You can now say that



        $$begin{align*}mathscr{F}^{-1}left{dfrac{1}{(1+iomega)^2}right} &= mathscr{F}^{-1}left{ idfrac{d}{domega} dfrac{1}{1+iomega}right}\
        \
        &= xmathscr{F}^{-1}left{ dfrac{1}{1+iomega}right}\
        \
        &= sqrt{2pi}xmathscr{F}^{-1}left{dfrac{1}{sqrt{2pi}}cdot dfrac{1}{1+iomega}right}\
        \
        &= sqrt{2pi}xe^{-x}H(x)\
        end{align*}$$






        share|cite|improve this answer









        $endgroup$



        A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.



        Observe that



        $$dfrac{1}{(1+iomega)^2} = idfrac{d}{domega} dfrac{1}{1+iomega}$$



        and note that the Fourier Transform has a derivative theorem



        $$mathscr{F}left{x^n f(x)right}=mathscr{F}left{x^n mathscr{F}^{-1}left{ F(omega)right}right}= i^ndfrac{d^n}{domega^n} F(omega)$$



        A table lookup of Fourier Transforms yields



        $$mathscr{F}left{e^{-ax}H(x)right} = dfrac{1}{sqrt{2pi}}cdotdfrac{1}{a+iomega}$$



        where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)



        You can now say that



        $$begin{align*}mathscr{F}^{-1}left{dfrac{1}{(1+iomega)^2}right} &= mathscr{F}^{-1}left{ idfrac{d}{domega} dfrac{1}{1+iomega}right}\
        \
        &= xmathscr{F}^{-1}left{ dfrac{1}{1+iomega}right}\
        \
        &= sqrt{2pi}xmathscr{F}^{-1}left{dfrac{1}{sqrt{2pi}}cdot dfrac{1}{1+iomega}right}\
        \
        &= sqrt{2pi}xe^{-x}H(x)\
        end{align*}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 23:11









        Andy WallsAndy Walls

        1,594128




        1,594128






























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