Mixing
![Does this series converges to some constant? [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Fourier transform of Gaussian is equal to derivative of Fourier transform of Gaussian times constant
Clash Royale CLAN TAG#URR8PPP
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I'm working on a few problems in Reed & Simon, and I ran across this problem.
Compute the Fourier transform of $f(x) = e^-alpha x^2/2$ via the following steps.
(a) Prove that $-lambda hatf(lambda)= alpha fracddlambdahatf(lambda)$ and conclude that $hatf(lambda) = ce^-lambda^2/(2alpha)$.
This is what I have so far (with help from first comment):
beginalign*
-fracalphalambda fracddlambda hatf(lambda) &= -fracalphalambda sqrt2pi int fracddlambda e^-ilambda xe^-alpha x^2/2 dx \
&= fracalpha ilambdasqrt2pi int x e^-alpha x^2/2 e^frac2ilambdaalpha dx \
&= fracalpha ilambdasqrt2pi Big[-fracilambdaalpha e^-fraclambda^22alphasqrtfrac2alphapiBig] \
&=frac1sqrtalphapie^-fraclambda^22alpha
endalign*
I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.
Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^-x^2/2$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.
normal-distribution fourier-transform
asked Jul 28 at 0:37
mathishard.butweloveit
1069
add a comment |Â
up vote
1
down vote
favorite
I'm working on a few problems in Reed & Simon, and I ran across this problem.
Compute the Fourier transform of $f(x) = e^-alpha x^2/2$ via the following steps.
(a) Prove that $-lambda hatf(lambda)= alpha fracddlambdahatf(lambda)$ and conclude that $hatf(lambda) = ce^-lambda^2/(2alpha)$.
This is what I have so far (with help from first comment):
beginalign*
-fracalphalambda fracddlambda hatf(lambda) &= -fracalphalambda sqrt2pi int fracddlambda e^-ilambda xe^-alpha x^2/2 dx \
&= fracalpha ilambdasqrt2pi int x e^-alpha x^2/2 e^frac2ilambdaalpha dx \
&= fracalpha ilambdasqrt2pi Big[-fracilambdaalpha e^-fraclambda^22alphasqrtfrac2alphapiBig] \
&=frac1sqrtalphapie^-fraclambda^22alpha
endalign*
I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.
Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^-x^2/2$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.
normal-distribution fourier-transform
asked Jul 28 at 0:37
mathishard.butweloveit
1069
2
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
2
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
2
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on a few problems in Reed & Simon, and I ran across this problem.
Compute the Fourier transform of $f(x) = e^-alpha x^2/2$ via the following steps.
(a) Prove that $-lambda hatf(lambda)= alpha fracddlambdahatf(lambda)$ and conclude that $hatf(lambda) = ce^-lambda^2/(2alpha)$.
This is what I have so far (with help from first comment):
beginalign*
-fracalphalambda fracddlambda hatf(lambda) &= -fracalphalambda sqrt2pi int fracddlambda e^-ilambda xe^-alpha x^2/2 dx \
&= fracalpha ilambdasqrt2pi int x e^-alpha x^2/2 e^frac2ilambdaalpha dx \
&= fracalpha ilambdasqrt2pi Big[-fracilambdaalpha e^-fraclambda^22alphasqrtfrac2alphapiBig] \
&=frac1sqrtalphapie^-fraclambda^22alpha
endalign*
I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.
Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^-x^2/2$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.
normal-distribution fourier-transform
asked Jul 28 at 0:37
mathishard.butweloveit
1069
I'm working on a few problems in Reed & Simon, and I ran across this problem.
Compute the Fourier transform of $f(x) = e^-alpha x^2/2$ via the following steps.
(a) Prove that $-lambda hatf(lambda)= alpha fracddlambdahatf(lambda)$ and conclude that $hatf(lambda) = ce^-lambda^2/(2alpha)$.
This is what I have so far (with help from first comment):
beginalign*
-fracalphalambda fracddlambda hatf(lambda) &= -fracalphalambda sqrt2pi int fracddlambda e^-ilambda xe^-alpha x^2/2 dx \
&= fracalpha ilambdasqrt2pi int x e^-alpha x^2/2 e^frac2ilambdaalpha dx \
&= fracalpha ilambdasqrt2pi Big[-fracilambdaalpha e^-fraclambda^22alphasqrtfrac2alphapiBig] \
&=frac1sqrtalphapie^-fraclambda^22alpha
endalign*
I'm not entirely sure where to go from here. I am going to keep thinking about it, but any hints would be appreciated.
Edits: I fixed the computations, all I have left to figure out on my own is the second portion of the problem, which probably isn't too bad using $u$-substitution and the fact that $e^-x^2/2$ is its own Fourier transform. The accepted answer provides a different and interesting way of looking at this problem as well.
normal-distribution fourier-transform
asked Jul 28 at 0:37
mathishard.butweloveit
1069
edited Jul 28 at 1:24
asked Jul 28 at 0:37
mathishard.butweloveit
1069
asked Jul 28 at 0:37
mathishard.butweloveit
1069
asked Jul 28 at 0:37
mathishard.butweloveit
1069
1069
2
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
2
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
2
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16
add a comment |Â
2
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
2
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
2
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16
2
2
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
2
2
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
2
2
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Theorem: The characteristic function of a standard normal random variable $X$ is
$
varphi_X(t)=e^-t^2/2
$.
Proof. Note that
beginalign*
varphi_X(t)=mathbbEleft[e^itXright] & =frac1sqrt2piint_mathbbRe^-fracx^22e^itxdx\
& =frac1sqrt2pileft(int_-infty^0e^-fracx^22e^itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac1sqrt2pileft(int_0^inftye^-fracx^22e^-itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac2sqrt2piint_0^inftye^-fracx^22cos(tx)dx.
endalign*
Now, take the derivative with respect to $t$ to get
$$
varphi_X^prime(t)=-frac2sqrt2piint_0^inftye^-fracx^22xsin(tx)dx.
$$
Integrate by parts to get
beginalign*
varphi_X^prime(t) & =frac2sqrt2pileft(e^-fracx^22sin(tx)mid_0^infty-tint_0^inftye^-fracx^22cos(tx)dxright)\
& =-frac2tsqrt2piint_0^inftye^-fracx^22cos(tx)dx\
& =-tvarphi_X(t).
endalign*
Note that
$$
varphi_X^prime(t)=-tvarphi_X(t)
$$
is an ODE with solution
$$
varphi_X(t)=ce^-t^2/2.
$$
We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $varphi_X(0)=1$ so that $c=1$, as desired.
answered Jul 28 at 0:49
parsiad
15.9k32253
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Theorem: The characteristic function of a standard normal random variable $X$ is
$
varphi_X(t)=e^-t^2/2
$.
Proof. Note that
beginalign*
varphi_X(t)=mathbbEleft[e^itXright] & =frac1sqrt2piint_mathbbRe^-fracx^22e^itxdx\
& =frac1sqrt2pileft(int_-infty^0e^-fracx^22e^itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac1sqrt2pileft(int_0^inftye^-fracx^22e^-itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac2sqrt2piint_0^inftye^-fracx^22cos(tx)dx.
endalign*
Now, take the derivative with respect to $t$ to get
$$
varphi_X^prime(t)=-frac2sqrt2piint_0^inftye^-fracx^22xsin(tx)dx.
$$
Integrate by parts to get
beginalign*
varphi_X^prime(t) & =frac2sqrt2pileft(e^-fracx^22sin(tx)mid_0^infty-tint_0^inftye^-fracx^22cos(tx)dxright)\
& =-frac2tsqrt2piint_0^inftye^-fracx^22cos(tx)dx\
& =-tvarphi_X(t).
endalign*
Note that
$$
varphi_X^prime(t)=-tvarphi_X(t)
$$
is an ODE with solution
$$
varphi_X(t)=ce^-t^2/2.
$$
We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $varphi_X(0)=1$ so that $c=1$, as desired.
answered Jul 28 at 0:49
parsiad
15.9k32253
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
add a comment |Â
up vote
2
down vote
accepted
Theorem: The characteristic function of a standard normal random variable $X$ is
$
varphi_X(t)=e^-t^2/2
$.
Proof. Note that
beginalign*
varphi_X(t)=mathbbEleft[e^itXright] & =frac1sqrt2piint_mathbbRe^-fracx^22e^itxdx\
& =frac1sqrt2pileft(int_-infty^0e^-fracx^22e^itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac1sqrt2pileft(int_0^inftye^-fracx^22e^-itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac2sqrt2piint_0^inftye^-fracx^22cos(tx)dx.
endalign*
Now, take the derivative with respect to $t$ to get
$$
varphi_X^prime(t)=-frac2sqrt2piint_0^inftye^-fracx^22xsin(tx)dx.
$$
Integrate by parts to get
beginalign*
varphi_X^prime(t) & =frac2sqrt2pileft(e^-fracx^22sin(tx)mid_0^infty-tint_0^inftye^-fracx^22cos(tx)dxright)\
& =-frac2tsqrt2piint_0^inftye^-fracx^22cos(tx)dx\
& =-tvarphi_X(t).
endalign*
Note that
$$
varphi_X^prime(t)=-tvarphi_X(t)
$$
is an ODE with solution
$$
varphi_X(t)=ce^-t^2/2.
$$
We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $varphi_X(0)=1$ so that $c=1$, as desired.
answered Jul 28 at 0:49
parsiad
15.9k32253
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Theorem: The characteristic function of a standard normal random variable $X$ is
$
varphi_X(t)=e^-t^2/2
$.
Proof. Note that
beginalign*
varphi_X(t)=mathbbEleft[e^itXright] & =frac1sqrt2piint_mathbbRe^-fracx^22e^itxdx\
& =frac1sqrt2pileft(int_-infty^0e^-fracx^22e^itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac1sqrt2pileft(int_0^inftye^-fracx^22e^-itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac2sqrt2piint_0^inftye^-fracx^22cos(tx)dx.
endalign*
Now, take the derivative with respect to $t$ to get
$$
varphi_X^prime(t)=-frac2sqrt2piint_0^inftye^-fracx^22xsin(tx)dx.
$$
Integrate by parts to get
beginalign*
varphi_X^prime(t) & =frac2sqrt2pileft(e^-fracx^22sin(tx)mid_0^infty-tint_0^inftye^-fracx^22cos(tx)dxright)\
& =-frac2tsqrt2piint_0^inftye^-fracx^22cos(tx)dx\
& =-tvarphi_X(t).
endalign*
Note that
$$
varphi_X^prime(t)=-tvarphi_X(t)
$$
is an ODE with solution
$$
varphi_X(t)=ce^-t^2/2.
$$
We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $varphi_X(0)=1$ so that $c=1$, as desired.
answered Jul 28 at 0:49
parsiad
15.9k32253
Theorem: The characteristic function of a standard normal random variable $X$ is
$
varphi_X(t)=e^-t^2/2
$.
Proof. Note that
beginalign*
varphi_X(t)=mathbbEleft[e^itXright] & =frac1sqrt2piint_mathbbRe^-fracx^22e^itxdx\
& =frac1sqrt2pileft(int_-infty^0e^-fracx^22e^itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac1sqrt2pileft(int_0^inftye^-fracx^22e^-itxdx+int_0^inftye^-fracx^22e^itxdxright)\
& =frac2sqrt2piint_0^inftye^-fracx^22cos(tx)dx.
endalign*
Now, take the derivative with respect to $t$ to get
$$
varphi_X^prime(t)=-frac2sqrt2piint_0^inftye^-fracx^22xsin(tx)dx.
$$
Integrate by parts to get
beginalign*
varphi_X^prime(t) & =frac2sqrt2pileft(e^-fracx^22sin(tx)mid_0^infty-tint_0^inftye^-fracx^22cos(tx)dxright)\
& =-frac2tsqrt2piint_0^inftye^-fracx^22cos(tx)dx\
& =-tvarphi_X(t).
endalign*
Note that
$$
varphi_X^prime(t)=-tvarphi_X(t)
$$
is an ODE with solution
$$
varphi_X(t)=ce^-t^2/2.
$$
We can figure out the value of the constant $c$ by recalling that the characteristic function has to satisfy $varphi_X(0)=1$ so that $c=1$, as desired.
answered Jul 28 at 0:49
parsiad
15.9k32253
answered Jul 28 at 0:49
parsiad
15.9k32253
answered Jul 28 at 0:49
parsiad
15.9k32253
answered Jul 28 at 0:49
parsiad
15.9k32253
15.9k32253
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
add a comment |Â
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
This is a very interesting proof. I don't know much about the standard random variable but I follow. Thank you.
– mathishard.butweloveit
Jul 28 at 1:11
1
1
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
Let $f$ be a probability density and $X$ be a random variable admitting that density. Then, the characteristic function $varphi_X$ is the Fourier transform of $f$ simply by definition. You can basically ignore this fact and just look at the integral, which you should recognize as the Fourier transform (though, you might have another constant factor depending on your definition). If you are satisfied with the response, feel free to accept.
– parsiad
Jul 28 at 1:16
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
I like the fact that this gives me a different angle, but I also edited the problem from the angle of the first comment. thanks
– mathishard.butweloveit
Jul 28 at 1:22
add a comment |Â
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2
Try integrating $int e^i lambda x (x e^-alpha x^2/2) ,mathrmdx$ by parts.
– stochasticboy321
Jul 28 at 0:48
2
Just for completeness' sake, I meant something a little different from what OP ended up doing with the hint: $$ int_mathbbR e^-ilambda x (x e^-alpha x^2/2) ,mathrmdx = left. e^-i lambda x frac- e^-alpha x^2 /2alpharight|_-infty^+ infty - i fraclambdaalphaint_mathbbR e^-ilambda x e^-alpha x^2/2 ,mathrmdx = -i frac lambda sqrt2pialpha hatf(lambda).$$
– stochasticboy321
Jul 28 at 2:16
2
So, from the second equality in the question, $$ -fracalphalambda fracmathrmd mathrmd lambda hatf (lambda) = fraci alphalambda sqrt2pi cdot -i frac lambda sqrt2pialpha hatf(lambda) = hatf (lambda),$$ thus showing the ODE required. Now one can invoke the conclusion of parsiad's argument.
– stochasticboy321
Jul 28 at 2:16