Fourier Transform of Heaviside Step Function












2












$begingroup$


Is it possible using the following property of Fourier transforms;



$$frac{d^nf(x)}{dx^n} = 2pi inu hat{f}(x)$$



to show that the Fourier transform of the heaviside step function is equal to;



$$hat{f}[theta(t)] = frac{1}{2pi i nu}$$



Using the property listed above we can write:



$$hat{f}[theta'(t)] = 2pi inuhat{f}[theta(t)]$$ But since we know the (distributional) derivative of the Heaviside step function is the Dirac delta function.



$$ hat{f}[delta(t)] = 2pi inuhat{f}[theta(t)]$$



$$implies hat{f}[theta(t)] = frac{hat{f}[delta(t)]}{2pi i nu}$$



And since the Fourier transform of the delta function is just $1$ we get:



$$ hat{f}[theta(t)] = frac{1}{2 pi i nu}$$



I am unsure if this process is correct or rigorous since my knowledge and understanding of distributions and generalised functions is limited.



If this method presented above is indeed allowed I am unsure why this answer contradicts the answer shown on Mathworld which states the Fourier transform of the Heaviside step function is equal to:



$$ frac{1}{2}bigl[delta(k)space - frac{i}{pi k}bigr]$$



There is also another article here https://www.cs.uaf.edu/~bueler/M611heaviside.pdf which does not return my answer.



Any comments, corrections or answers are greatly appreciated.



Thanks!






fourier-analysis distribution-theory fourier-transform dirac-delta







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$endgroup$












  • $begingroup$
    $displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
    $endgroup$
    – Felix Marin
    May 3 '18 at 6:32


















2












$begingroup$


Is it possible using the following property of Fourier transforms;



$$frac{d^nf(x)}{dx^n} = 2pi inu hat{f}(x)$$



to show that the Fourier transform of the heaviside step function is equal to;



$$hat{f}[theta(t)] = frac{1}{2pi i nu}$$



Using the property listed above we can write:



$$hat{f}[theta'(t)] = 2pi inuhat{f}[theta(t)]$$ But since we know the (distributional) derivative of the Heaviside step function is the Dirac delta function.



$$ hat{f}[delta(t)] = 2pi inuhat{f}[theta(t)]$$



$$implies hat{f}[theta(t)] = frac{hat{f}[delta(t)]}{2pi i nu}$$



And since the Fourier transform of the delta function is just $1$ we get:



$$ hat{f}[theta(t)] = frac{1}{2 pi i nu}$$



I am unsure if this process is correct or rigorous since my knowledge and understanding of distributions and generalised functions is limited.



If this method presented above is indeed allowed I am unsure why this answer contradicts the answer shown on Mathworld which states the Fourier transform of the Heaviside step function is equal to:



$$ frac{1}{2}bigl[delta(k)space - frac{i}{pi k}bigr]$$



There is also another article here https://www.cs.uaf.edu/~bueler/M611heaviside.pdf which does not return my answer.



Any comments, corrections or answers are greatly appreciated.



Thanks!






fourier-analysis distribution-theory fourier-transform dirac-delta







share|cite|improve this question









$endgroup$












  • $begingroup$
    $displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
    $endgroup$
    – Felix Marin
    May 3 '18 at 6:32
















2












2








2





$begingroup$


Is it possible using the following property of Fourier transforms;



$$frac{d^nf(x)}{dx^n} = 2pi inu hat{f}(x)$$



to show that the Fourier transform of the heaviside step function is equal to;



$$hat{f}[theta(t)] = frac{1}{2pi i nu}$$



Using the property listed above we can write:



$$hat{f}[theta'(t)] = 2pi inuhat{f}[theta(t)]$$ But since we know the (distributional) derivative of the Heaviside step function is the Dirac delta function.



$$ hat{f}[delta(t)] = 2pi inuhat{f}[theta(t)]$$



$$implies hat{f}[theta(t)] = frac{hat{f}[delta(t)]}{2pi i nu}$$



And since the Fourier transform of the delta function is just $1$ we get:



$$ hat{f}[theta(t)] = frac{1}{2 pi i nu}$$



I am unsure if this process is correct or rigorous since my knowledge and understanding of distributions and generalised functions is limited.



If this method presented above is indeed allowed I am unsure why this answer contradicts the answer shown on Mathworld which states the Fourier transform of the Heaviside step function is equal to:



$$ frac{1}{2}bigl[delta(k)space - frac{i}{pi k}bigr]$$



There is also another article here https://www.cs.uaf.edu/~bueler/M611heaviside.pdf which does not return my answer.



Any comments, corrections or answers are greatly appreciated.



Thanks!






fourier-analysis distribution-theory fourier-transform dirac-delta







share|cite|improve this question









$endgroup$




Is it possible using the following property of Fourier transforms;



$$frac{d^nf(x)}{dx^n} = 2pi inu hat{f}(x)$$



to show that the Fourier transform of the heaviside step function is equal to;



$$hat{f}[theta(t)] = frac{1}{2pi i nu}$$



Using the property listed above we can write:



$$hat{f}[theta'(t)] = 2pi inuhat{f}[theta(t)]$$ But since we know the (distributional) derivative of the Heaviside step function is the Dirac delta function.



$$ hat{f}[delta(t)] = 2pi inuhat{f}[theta(t)]$$



$$implies hat{f}[theta(t)] = frac{hat{f}[delta(t)]}{2pi i nu}$$



And since the Fourier transform of the delta function is just $1$ we get:



$$ hat{f}[theta(t)] = frac{1}{2 pi i nu}$$



I am unsure if this process is correct or rigorous since my knowledge and understanding of distributions and generalised functions is limited.



If this method presented above is indeed allowed I am unsure why this answer contradicts the answer shown on Mathworld which states the Fourier transform of the Heaviside step function is equal to:



$$ frac{1}{2}bigl[delta(k)space - frac{i}{pi k}bigr]$$



There is also another article here https://www.cs.uaf.edu/~bueler/M611heaviside.pdf which does not return my answer.



Any comments, corrections or answers are greatly appreciated.



Thanks!






fourier-analysis distribution-theory fourier-transform dirac-delta




fourier-analysis distribution-theory fourier-transform dirac-delta






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share|cite|improve this question











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asked May 3 '18 at 5:45









LegendaryLegendary

327




327












  • $begingroup$
    $displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
    $endgroup$
    – Felix Marin
    May 3 '18 at 6:32




















  • $begingroup$
    $displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
    $endgroup$
    – Felix Marin
    May 3 '18 at 6:32


















$begingroup$
$displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
$endgroup$
– Felix Marin
May 3 '18 at 6:32






$begingroup$
$displaystylemathrm{H}left(xright) = int_{-infty}^{infty}{mathrm{e}^{mathrm{i}kx} over k - mathrm{i}0^{+}},{mathrm{d}k over 2pimathrm{i}}$ where $displaystyle x in mathbb{R}setminusleft{0right}$
$endgroup$
– Felix Marin
May 3 '18 at 6:32












1 Answer
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$begingroup$

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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
mrm{H}pars{x} & = int_{-infty}^{infty}{expo{ic kx} over k - ic 0^{+}},
{dd k over 2piic}
\[5mm]
hat{mrm{H}}pars{k} & = {1 over k -ic 0^{+}},{1 over ic} =
-icbracks{mrm{P.V.}{1 over k} + icpi,deltapars{k}} =
-ic,mrm{P.V.}{1 over k} + pi,deltapars{k}
end{align}






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
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    begin{align}
    mrm{H}pars{x} & = int_{-infty}^{infty}{expo{ic kx} over k - ic 0^{+}},
    {dd k over 2piic}
    \[5mm]
    hat{mrm{H}}pars{k} & = {1 over k -ic 0^{+}},{1 over ic} =
    -icbracks{mrm{P.V.}{1 over k} + icpi,deltapars{k}} =
    -ic,mrm{P.V.}{1 over k} + pi,deltapars{k}
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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      begin{align}
      mrm{H}pars{x} & = int_{-infty}^{infty}{expo{ic kx} over k - ic 0^{+}},
      {dd k over 2piic}
      \[5mm]
      hat{mrm{H}}pars{k} & = {1 over k -ic 0^{+}},{1 over ic} =
      -icbracks{mrm{P.V.}{1 over k} + icpi,deltapars{k}} =
      -ic,mrm{P.V.}{1 over k} + pi,deltapars{k}
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
        newcommand{ds}[1]{displaystyle{#1}}
        newcommand{expo}[1]{,mathrm{e}^{#1},}
        newcommand{ic}{mathrm{i}}
        newcommand{mc}[1]{mathcal{#1}}
        newcommand{mrm}[1]{mathrm{#1}}
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        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        mrm{H}pars{x} & = int_{-infty}^{infty}{expo{ic kx} over k - ic 0^{+}},
        {dd k over 2piic}
        \[5mm]
        hat{mrm{H}}pars{k} & = {1 over k -ic 0^{+}},{1 over ic} =
        -icbracks{mrm{P.V.}{1 over k} + icpi,deltapars{k}} =
        -ic,mrm{P.V.}{1 over k} + pi,deltapars{k}
        end{align}






        share|cite|improve this answer









        $endgroup$



        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
        newcommand{dd}{mathrm{d}}
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        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
        newcommand{verts}[1]{leftvert,{#1},rightvert}$
        begin{align}
        mrm{H}pars{x} & = int_{-infty}^{infty}{expo{ic kx} over k - ic 0^{+}},
        {dd k over 2piic}
        \[5mm]
        hat{mrm{H}}pars{k} & = {1 over k -ic 0^{+}},{1 over ic} =
        -icbracks{mrm{P.V.}{1 over k} + icpi,deltapars{k}} =
        -ic,mrm{P.V.}{1 over k} + pi,deltapars{k}
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 3 '18 at 6:39









        Felix MarinFelix Marin

        68.1k7109143




        68.1k7109143






























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