Derivative of Fourier transform using residue theorem
$begingroup$
If we define the fourier transform of f as
$$hat{f}(omega) = frac{1}{sqrt{2}}int_{-infty}^infty f(x) e^{-iomega x} dx$$
then if f is differentiable, and the integrals for $hat{f}$ and $hat{f'}$ both converge then
$$hat{f'}(w) = frac{1}{iomega}hat{f}(omega)$$
My first attempt was trying to use the fact that since both $hat{f}$ and $hat{f'}$ converge, then the improper integral in their respective definitions can be rewritten using the residue theorem, but this is as far as I've gotten and I am stuck. Someone suggested I try using integration by parts but I am unsure where to use this.
complex-analysis fourier-transform residue-calculus
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
$endgroup$
add a comment |
$begingroup$
If we define the fourier transform of f as
$$hat{f}(omega) = frac{1}{sqrt{2}}int_{-infty}^infty f(x) e^{-iomega x} dx$$
then if f is differentiable, and the integrals for $hat{f}$ and $hat{f'}$ both converge then
$$hat{f'}(w) = frac{1}{iomega}hat{f}(omega)$$
My first attempt was trying to use the fact that since both $hat{f}$ and $hat{f'}$ converge, then the improper integral in their respective definitions can be rewritten using the residue theorem, but this is as far as I've gotten and I am stuck. Someone suggested I try using integration by parts but I am unsure where to use this.
complex-analysis fourier-transform residue-calculus
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
$endgroup$
$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22
add a comment |
$begingroup$
If we define the fourier transform of f as
$$hat{f}(omega) = frac{1}{sqrt{2}}int_{-infty}^infty f(x) e^{-iomega x} dx$$
then if f is differentiable, and the integrals for $hat{f}$ and $hat{f'}$ both converge then
$$hat{f'}(w) = frac{1}{iomega}hat{f}(omega)$$
My first attempt was trying to use the fact that since both $hat{f}$ and $hat{f'}$ converge, then the improper integral in their respective definitions can be rewritten using the residue theorem, but this is as far as I've gotten and I am stuck. Someone suggested I try using integration by parts but I am unsure where to use this.
complex-analysis fourier-transform residue-calculus
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
$endgroup$
If we define the fourier transform of f as
$$hat{f}(omega) = frac{1}{sqrt{2}}int_{-infty}^infty f(x) e^{-iomega x} dx$$
then if f is differentiable, and the integrals for $hat{f}$ and $hat{f'}$ both converge then
$$hat{f'}(w) = frac{1}{iomega}hat{f}(omega)$$
My first attempt was trying to use the fact that since both $hat{f}$ and $hat{f'}$ converge, then the improper integral in their respective definitions can be rewritten using the residue theorem, but this is as far as I've gotten and I am stuck. Someone suggested I try using integration by parts but I am unsure where to use this.
complex-analysis fourier-transform residue-calculus
complex-analysis fourier-transform residue-calculus
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
asked Dec 4 '18 at 6:52
Richard VillalobosRichard Villalobos
1567
1567
$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22
add a comment |
$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22
$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22
$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22
add a comment |
1 Answer
1
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oldest
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$begingroup$
begin{eqnarray}
int_{-infty}^{+infty} frac{{rm d}f(x)}{{rm d}x} e^{-iomega x}~{rm d}x &=& int_{-infty}^{+infty}left[frac{{rm d}}{{rm d}x}(f(x)e^{-iomega x}) - f(x)frac{{rm d}e^{-iomega x}}{{rm d}x} right]{rm d}x \
&=& left.f(x)e^{-iomega x}right|_{-infty}^{+infty} + iomega int_{-infty}^{+infty}f(x)e^{-iomega x}~{rm d}x
end{eqnarray}
The first term vanishes if the function $f$ goes to 0 at infinity. From here
$$
hat{f'}(omega) = (iomega) hat{f}(omega)
$$
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
begin{eqnarray}
int_{-infty}^{+infty} frac{{rm d}f(x)}{{rm d}x} e^{-iomega x}~{rm d}x &=& int_{-infty}^{+infty}left[frac{{rm d}}{{rm d}x}(f(x)e^{-iomega x}) - f(x)frac{{rm d}e^{-iomega x}}{{rm d}x} right]{rm d}x \
&=& left.f(x)e^{-iomega x}right|_{-infty}^{+infty} + iomega int_{-infty}^{+infty}f(x)e^{-iomega x}~{rm d}x
end{eqnarray}
The first term vanishes if the function $f$ goes to 0 at infinity. From here
$$
hat{f'}(omega) = (iomega) hat{f}(omega)
$$
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
begin{eqnarray}
int_{-infty}^{+infty} frac{{rm d}f(x)}{{rm d}x} e^{-iomega x}~{rm d}x &=& int_{-infty}^{+infty}left[frac{{rm d}}{{rm d}x}(f(x)e^{-iomega x}) - f(x)frac{{rm d}e^{-iomega x}}{{rm d}x} right]{rm d}x \
&=& left.f(x)e^{-iomega x}right|_{-infty}^{+infty} + iomega int_{-infty}^{+infty}f(x)e^{-iomega x}~{rm d}x
end{eqnarray}
The first term vanishes if the function $f$ goes to 0 at infinity. From here
$$
hat{f'}(omega) = (iomega) hat{f}(omega)
$$
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
begin{eqnarray}
int_{-infty}^{+infty} frac{{rm d}f(x)}{{rm d}x} e^{-iomega x}~{rm d}x &=& int_{-infty}^{+infty}left[frac{{rm d}}{{rm d}x}(f(x)e^{-iomega x}) - f(x)frac{{rm d}e^{-iomega x}}{{rm d}x} right]{rm d}x \
&=& left.f(x)e^{-iomega x}right|_{-infty}^{+infty} + iomega int_{-infty}^{+infty}f(x)e^{-iomega x}~{rm d}x
end{eqnarray}
The first term vanishes if the function $f$ goes to 0 at infinity. From here
$$
hat{f'}(omega) = (iomega) hat{f}(omega)
$$
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
$endgroup$
begin{eqnarray}
int_{-infty}^{+infty} frac{{rm d}f(x)}{{rm d}x} e^{-iomega x}~{rm d}x &=& int_{-infty}^{+infty}left[frac{{rm d}}{{rm d}x}(f(x)e^{-iomega x}) - f(x)frac{{rm d}e^{-iomega x}}{{rm d}x} right]{rm d}x \
&=& left.f(x)e^{-iomega x}right|_{-infty}^{+infty} + iomega int_{-infty}^{+infty}f(x)e^{-iomega x}~{rm d}x
end{eqnarray}
The first term vanishes if the function $f$ goes to 0 at infinity. From here
$$
hat{f'}(omega) = (iomega) hat{f}(omega)
$$
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
answered Dec 4 '18 at 13:22
caveraccaverac
14.5k31130
14.5k31130
add a comment |
add a comment |
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$begingroup$
The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $frac1{sqrt{2pi}}$
$endgroup$
– reuns
Dec 4 '18 at 8:22