Value of an integral depending on a function and its cosine transform
$begingroup$
Let's consider $g(x)$ a function in $C^{infty}(mathbb{R}^+ to mathbb{C})$ such that $g(x)$ is asymptotic to $x^{alpha}$ for $x$ near $+infty$ with $g(0)=0$, and $alpha$ a complex number such that $Re(alpha)<-frac{1}{2}$.
If $g$ satisfy following condition:
$$intlimits_{0}^{infty} g(x) overline{x^{alpha} } dx=0 $$
then following integral is well defined:
$$ I=intlimits_{0}^infty frac{1}{x} intlimits_{x}^{infty} g(y) ;overline{y^{alpha} }+4 ; hat{g(y)} ;overline{hat{y^{alpha} }} ; dy$$
where $hat{f}(x)$ is the cosine transform ($hat{f}(x)=intlimits_{0}^infty f(t) cos(2pi xt) dt)$
It is quite intuitive that $I$ is not zero for all $g$ functions satisfying given conditions above, but how to prove it ?
How to prove that there exists $g$ functions such that $I ne 0$ ? Is there an elegant / simple way ? Which functions can be used for a simple proof ? I took linear combinations of functions of the form $x^b e^{-x^2}$ but calculation of $I$ is not so simple ! What type of functions can be good candidates for simple demonstration ? Any trick to show there is a contradiction if we assume all $g$ funcitons give $I=0$ ?
real-analysis integration complex-analysis fourier-analysis
asked Jan 3 at 11:41
BertrandBertrand
1726
$endgroup$
add a comment |
$begingroup$
Let's consider $g(x)$ a function in $C^{infty}(mathbb{R}^+ to mathbb{C})$ such that $g(x)$ is asymptotic to $x^{alpha}$ for $x$ near $+infty$ with $g(0)=0$, and $alpha$ a complex number such that $Re(alpha)<-frac{1}{2}$.
If $g$ satisfy following condition:
$$intlimits_{0}^{infty} g(x) overline{x^{alpha} } dx=0 $$
then following integral is well defined:
$$ I=intlimits_{0}^infty frac{1}{x} intlimits_{x}^{infty} g(y) ;overline{y^{alpha} }+4 ; hat{g(y)} ;overline{hat{y^{alpha} }} ; dy$$
where $hat{f}(x)$ is the cosine transform ($hat{f}(x)=intlimits_{0}^infty f(t) cos(2pi xt) dt)$
It is quite intuitive that $I$ is not zero for all $g$ functions satisfying given conditions above, but how to prove it ?
How to prove that there exists $g$ functions such that $I ne 0$ ? Is there an elegant / simple way ? Which functions can be used for a simple proof ? I took linear combinations of functions of the form $x^b e^{-x^2}$ but calculation of $I$ is not so simple ! What type of functions can be good candidates for simple demonstration ? Any trick to show there is a contradiction if we assume all $g$ funcitons give $I=0$ ?
real-analysis integration complex-analysis fourier-analysis
asked Jan 3 at 11:41
BertrandBertrand
1726
$endgroup$
add a comment |
$begingroup$
Let's consider $g(x)$ a function in $C^{infty}(mathbb{R}^+ to mathbb{C})$ such that $g(x)$ is asymptotic to $x^{alpha}$ for $x$ near $+infty$ with $g(0)=0$, and $alpha$ a complex number such that $Re(alpha)<-frac{1}{2}$.
If $g$ satisfy following condition:
$$intlimits_{0}^{infty} g(x) overline{x^{alpha} } dx=0 $$
then following integral is well defined:
$$ I=intlimits_{0}^infty frac{1}{x} intlimits_{x}^{infty} g(y) ;overline{y^{alpha} }+4 ; hat{g(y)} ;overline{hat{y^{alpha} }} ; dy$$
where $hat{f}(x)$ is the cosine transform ($hat{f}(x)=intlimits_{0}^infty f(t) cos(2pi xt) dt)$
It is quite intuitive that $I$ is not zero for all $g$ functions satisfying given conditions above, but how to prove it ?
How to prove that there exists $g$ functions such that $I ne 0$ ? Is there an elegant / simple way ? Which functions can be used for a simple proof ? I took linear combinations of functions of the form $x^b e^{-x^2}$ but calculation of $I$ is not so simple ! What type of functions can be good candidates for simple demonstration ? Any trick to show there is a contradiction if we assume all $g$ funcitons give $I=0$ ?
real-analysis integration complex-analysis fourier-analysis
asked Jan 3 at 11:41
BertrandBertrand
1726
$endgroup$
Let's consider $g(x)$ a function in $C^{infty}(mathbb{R}^+ to mathbb{C})$ such that $g(x)$ is asymptotic to $x^{alpha}$ for $x$ near $+infty$ with $g(0)=0$, and $alpha$ a complex number such that $Re(alpha)<-frac{1}{2}$.
If $g$ satisfy following condition:
$$intlimits_{0}^{infty} g(x) overline{x^{alpha} } dx=0 $$
then following integral is well defined:
$$ I=intlimits_{0}^infty frac{1}{x} intlimits_{x}^{infty} g(y) ;overline{y^{alpha} }+4 ; hat{g(y)} ;overline{hat{y^{alpha} }} ; dy$$
where $hat{f}(x)$ is the cosine transform ($hat{f}(x)=intlimits_{0}^infty f(t) cos(2pi xt) dt)$
It is quite intuitive that $I$ is not zero for all $g$ functions satisfying given conditions above, but how to prove it ?
How to prove that there exists $g$ functions such that $I ne 0$ ? Is there an elegant / simple way ? Which functions can be used for a simple proof ? I took linear combinations of functions of the form $x^b e^{-x^2}$ but calculation of $I$ is not so simple ! What type of functions can be good candidates for simple demonstration ? Any trick to show there is a contradiction if we assume all $g$ funcitons give $I=0$ ?
real-analysis integration complex-analysis fourier-analysis
real-analysis integration complex-analysis fourier-analysis
asked Jan 3 at 11:41
BertrandBertrand
1726
asked Jan 3 at 11:41
BertrandBertrand
1726
edited Jan 3 at 12:18
Bertrand
asked Jan 3 at 11:41
BertrandBertrand
1726
asked Jan 3 at 11:41
BertrandBertrand
1726
asked Jan 3 at 11:41
BertrandBertrand
1726
1726
add a comment |
add a comment |
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