Inverse Fourier transformation of $frac{1}{(1+w^2)^2}$?
$begingroup$
I have tried to crack this via basic definition, but I am unable to solve the integral:
$$int_{-infty} ^{infty} frac{1}{(1+w^2)^2} e^{-iwx} dw $$
Kindly guide me a bit.
Thanks in anticipation.
fourier-analysis fourier-transform
edited Dec 19 '18 at 7:09
dmtri
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
$endgroup$
add a comment |
$begingroup$
I have tried to crack this via basic definition, but I am unable to solve the integral:
$$int_{-infty} ^{infty} frac{1}{(1+w^2)^2} e^{-iwx} dw $$
Kindly guide me a bit.
Thanks in anticipation.
fourier-analysis fourier-transform
edited Dec 19 '18 at 7:09
dmtri
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
$endgroup$
add a comment |
$begingroup$
I have tried to crack this via basic definition, but I am unable to solve the integral:
$$int_{-infty} ^{infty} frac{1}{(1+w^2)^2} e^{-iwx} dw $$
Kindly guide me a bit.
Thanks in anticipation.
fourier-analysis fourier-transform
edited Dec 19 '18 at 7:09
dmtri
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
$endgroup$
I have tried to crack this via basic definition, but I am unable to solve the integral:
$$int_{-infty} ^{infty} frac{1}{(1+w^2)^2} e^{-iwx} dw $$
Kindly guide me a bit.
Thanks in anticipation.
fourier-analysis fourier-transform
fourier-analysis fourier-transform
edited Dec 19 '18 at 7:09
dmtri
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
edited Dec 19 '18 at 7:09
dmtri
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
edited Dec 19 '18 at 7:09
dmtri
1,7632521
edited Dec 19 '18 at 7:09
dmtri
1,7632521
edited Dec 19 '18 at 7:09
dmtri
1,7632521
1,7632521
asked Dec 19 '18 at 6:36
CommandoCommando
133
asked Dec 19 '18 at 6:36
CommandoCommando
133
asked Dec 19 '18 at 6:36
CommandoCommando
133
133
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$
(which is $pi e^{-|x|}$) and recall that the Fourier transform converts multiplication
into convolution.
Therefore, for $g(x)=e^{-|x|}$,
$$int_{-infty}^inftyfrac{e^{-iwx},dw}{(1+w^2)^2}
=g*g(x)=int_{-infty}^infty g(y)g(x-y),dy
=pi^2int_{-infty}^inftyexp(-|y|-|x-y|),dy.$$
One can split this into three integrals over intervals where $-|y|-|x-y|$
is linear on each.
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$
(which is $pi e^{-|x|}$) and recall that the Fourier transform converts multiplication
into convolution.
Therefore, for $g(x)=e^{-|x|}$,
$$int_{-infty}^inftyfrac{e^{-iwx},dw}{(1+w^2)^2}
=g*g(x)=int_{-infty}^infty g(y)g(x-y),dy
=pi^2int_{-infty}^inftyexp(-|y|-|x-y|),dy.$$
One can split this into three integrals over intervals where $-|y|-|x-y|$
is linear on each.
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
add a comment |
$begingroup$
You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$
(which is $pi e^{-|x|}$) and recall that the Fourier transform converts multiplication
into convolution.
Therefore, for $g(x)=e^{-|x|}$,
$$int_{-infty}^inftyfrac{e^{-iwx},dw}{(1+w^2)^2}
=g*g(x)=int_{-infty}^infty g(y)g(x-y),dy
=pi^2int_{-infty}^inftyexp(-|y|-|x-y|),dy.$$
One can split this into three integrals over intervals where $-|y|-|x-y|$
is linear on each.
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
add a comment |
$begingroup$
You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$
(which is $pi e^{-|x|}$) and recall that the Fourier transform converts multiplication
into convolution.
Therefore, for $g(x)=e^{-|x|}$,
$$int_{-infty}^inftyfrac{e^{-iwx},dw}{(1+w^2)^2}
=g*g(x)=int_{-infty}^infty g(y)g(x-y),dy
=pi^2int_{-infty}^inftyexp(-|y|-|x-y|),dy.$$
One can split this into three integrals over intervals where $-|y|-|x-y|$
is linear on each.
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
You can avoid doing the integral if you know the Fourier transform of $1/(1+w^2)$
(which is $pi e^{-|x|}$) and recall that the Fourier transform converts multiplication
into convolution.
Therefore, for $g(x)=e^{-|x|}$,
$$int_{-infty}^inftyfrac{e^{-iwx},dw}{(1+w^2)^2}
=g*g(x)=int_{-infty}^infty g(y)g(x-y),dy
=pi^2int_{-infty}^inftyexp(-|y|-|x-y|),dy.$$
One can split this into three integrals over intervals where $-|y|-|x-y|$
is linear on each.
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Dec 19 '18 at 6:46
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
add a comment |
add a comment |
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