Fourier Transform of Entropic Utility Function
I'm trying find the Fourier transform $F(u)$ for the utility function for the entropic risk measure, which is given below. However my knowledge on Fourier analysis is fairly poor.
$f(x)=dfrac{e^{gamma x}-1}{gamma} $
I'm running into problems because when I try to do the integration I end up evaluating $dfrac{e^{x(gamma +iu)}}{gamma (gamma +iu)}-dfrac{e^{iux}}{iugamma}$ at infinity. Could anyone shed some light on what I'm messing up?
fourier-transform
asked Nov 21 '18 at 12:55
shrut9
204
add a comment |
I'm trying find the Fourier transform $F(u)$ for the utility function for the entropic risk measure, which is given below. However my knowledge on Fourier analysis is fairly poor.
$f(x)=dfrac{e^{gamma x}-1}{gamma} $
I'm running into problems because when I try to do the integration I end up evaluating $dfrac{e^{x(gamma +iu)}}{gamma (gamma +iu)}-dfrac{e^{iux}}{iugamma}$ at infinity. Could anyone shed some light on what I'm messing up?
fourier-transform
asked Nov 21 '18 at 12:55
shrut9
204
How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34
add a comment |
I'm trying find the Fourier transform $F(u)$ for the utility function for the entropic risk measure, which is given below. However my knowledge on Fourier analysis is fairly poor.
$f(x)=dfrac{e^{gamma x}-1}{gamma} $
I'm running into problems because when I try to do the integration I end up evaluating $dfrac{e^{x(gamma +iu)}}{gamma (gamma +iu)}-dfrac{e^{iux}}{iugamma}$ at infinity. Could anyone shed some light on what I'm messing up?
fourier-transform
asked Nov 21 '18 at 12:55
shrut9
204
I'm trying find the Fourier transform $F(u)$ for the utility function for the entropic risk measure, which is given below. However my knowledge on Fourier analysis is fairly poor.
$f(x)=dfrac{e^{gamma x}-1}{gamma} $
I'm running into problems because when I try to do the integration I end up evaluating $dfrac{e^{x(gamma +iu)}}{gamma (gamma +iu)}-dfrac{e^{iux}}{iugamma}$ at infinity. Could anyone shed some light on what I'm messing up?
fourier-transform
fourier-transform
asked Nov 21 '18 at 12:55
shrut9
204
asked Nov 21 '18 at 12:55
shrut9
204
asked Nov 21 '18 at 12:55
shrut9
204
asked Nov 21 '18 at 12:55
shrut9
204
asked Nov 21 '18 at 12:55
shrut9
204
204
How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34
add a comment |
How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34
How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34
add a comment |
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How do you compute the Fourier transform of $f$ when $f notin L^{1}(mathbb{R})$? Note that if $gamma > 0$ then $f$ blows up at $+infty$ and if $gamma < 0$ then $f$ blows up at $-infty$. Could you show us your working?
– Mattos
Nov 21 '18 at 13:08
$e^{gamma x}$ does not have a Fourier Transform even in a distributional sense. $e^{-|gamma x|}$ does have a Fourier Transform.
– Andy Walls
Nov 23 '18 at 2:34