Mixing
Characteristic Function of Gaussian/ Change of Variables in Complex Plane
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In the process of finding the characteristic function of a Guassian random variable, one must evaluate the following integral:
$$
int_-infty^infty e^-(x- it)^2 dx.
$$
One way to evaluate this is to use a contour integral over a rectangle and then use the residue theorem. This is presumably rigorous. Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line. Note that $dy/dx = 1$. My question is whether this method is rigorous or can be rigorously justified?
probability complex-analysis analysis
asked Jul 21 at 15:46
JohnKnoxV
828113
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up vote
1
down vote
favorite
In the process of finding the characteristic function of a Guassian random variable, one must evaluate the following integral:
$$
int_-infty^infty e^-(x- it)^2 dx.
$$
One way to evaluate this is to use a contour integral over a rectangle and then use the residue theorem. This is presumably rigorous. Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line. Note that $dy/dx = 1$. My question is whether this method is rigorous or can be rigorously justified?
probability complex-analysis analysis
asked Jul 21 at 15:46
JohnKnoxV
828113
The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the process of finding the characteristic function of a Guassian random variable, one must evaluate the following integral:
$$
int_-infty^infty e^-(x- it)^2 dx.
$$
One way to evaluate this is to use a contour integral over a rectangle and then use the residue theorem. This is presumably rigorous. Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line. Note that $dy/dx = 1$. My question is whether this method is rigorous or can be rigorously justified?
probability complex-analysis analysis
asked Jul 21 at 15:46
JohnKnoxV
828113
In the process of finding the characteristic function of a Guassian random variable, one must evaluate the following integral:
$$
int_-infty^infty e^-(x- it)^2 dx.
$$
One way to evaluate this is to use a contour integral over a rectangle and then use the residue theorem. This is presumably rigorous. Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line. Note that $dy/dx = 1$. My question is whether this method is rigorous or can be rigorously justified?
probability complex-analysis analysis
asked Jul 21 at 15:46
JohnKnoxV
828113
asked Jul 21 at 15:46
JohnKnoxV
828113
asked Jul 21 at 15:46
JohnKnoxV
828113
asked Jul 21 at 15:46
JohnKnoxV
828113
828113
The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08
add a comment |Â
The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08
The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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Here is one approach that does not require a change of variables and contour deformation. First we write
$$beginalign
int_-infty^infty e^-(x-it)^2,dx&=e^t^2int_-infty^infty e^-x^2+i2xt,dx\\
&=e^t^2int_-infty^infty e^-x^2cos(tx),dxtag1
endalign$$
Let $f(t)$ be given by the integral
$$f(t)=int_-infty^infty e^-x^2cos(tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_-infty^infty (-xe^-x^2)sin(tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(tx)$ and $v=frac12e^-x^2$ we obtain the ODE
$$f'(t)=-tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^-t^2tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_-infty^infty e^-(x-it)^2,dx=sqrt pi$$
answered Jul 21 at 18:05
Mark Viola
126k1172167
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
add a comment |Â
up vote
0
down vote
Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.
Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $operatornameIm(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $displaystyleoint_Gamma(R) e^-z^2, dz$, where $Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^-z^2$ along the vertical edges can be shown to be $O(e^-R^2)$ as $R to infty$. Since $e^-z^2$ is entire, Cauchy's theorem gives $displaystyleoint_Gamma(R) e^-z^2, dz = 0$. Hence $displaystylelim_Rto infty int_-R - it^R - it e^-z^2, dz = lim_Rto infty int_-R^R e^-x^2, dx$, or $$int_-infty^infty e^-(x-it)^2, dx = int_-infty^infty e^-x^2, dx$$A similar argument holds when $t < 0$.
answered Jul 22 at 20:45
kobe
33.9k22146
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is one approach that does not require a change of variables and contour deformation. First we write
$$beginalign
int_-infty^infty e^-(x-it)^2,dx&=e^t^2int_-infty^infty e^-x^2+i2xt,dx\\
&=e^t^2int_-infty^infty e^-x^2cos(tx),dxtag1
endalign$$
Let $f(t)$ be given by the integral
$$f(t)=int_-infty^infty e^-x^2cos(tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_-infty^infty (-xe^-x^2)sin(tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(tx)$ and $v=frac12e^-x^2$ we obtain the ODE
$$f'(t)=-tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^-t^2tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_-infty^infty e^-(x-it)^2,dx=sqrt pi$$
answered Jul 21 at 18:05
Mark Viola
126k1172167
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
add a comment |Â
up vote
1
down vote
Here is one approach that does not require a change of variables and contour deformation. First we write
$$beginalign
int_-infty^infty e^-(x-it)^2,dx&=e^t^2int_-infty^infty e^-x^2+i2xt,dx\\
&=e^t^2int_-infty^infty e^-x^2cos(tx),dxtag1
endalign$$
Let $f(t)$ be given by the integral
$$f(t)=int_-infty^infty e^-x^2cos(tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_-infty^infty (-xe^-x^2)sin(tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(tx)$ and $v=frac12e^-x^2$ we obtain the ODE
$$f'(t)=-tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^-t^2tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_-infty^infty e^-(x-it)^2,dx=sqrt pi$$
answered Jul 21 at 18:05
Mark Viola
126k1172167
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is one approach that does not require a change of variables and contour deformation. First we write
$$beginalign
int_-infty^infty e^-(x-it)^2,dx&=e^t^2int_-infty^infty e^-x^2+i2xt,dx\\
&=e^t^2int_-infty^infty e^-x^2cos(tx),dxtag1
endalign$$
Let $f(t)$ be given by the integral
$$f(t)=int_-infty^infty e^-x^2cos(tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_-infty^infty (-xe^-x^2)sin(tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(tx)$ and $v=frac12e^-x^2$ we obtain the ODE
$$f'(t)=-tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^-t^2tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_-infty^infty e^-(x-it)^2,dx=sqrt pi$$
answered Jul 21 at 18:05
Mark Viola
126k1172167
Here is one approach that does not require a change of variables and contour deformation. First we write
$$beginalign
int_-infty^infty e^-(x-it)^2,dx&=e^t^2int_-infty^infty e^-x^2+i2xt,dx\\
&=e^t^2int_-infty^infty e^-x^2cos(tx),dxtag1
endalign$$
Let $f(t)$ be given by the integral
$$f(t)=int_-infty^infty e^-x^2cos(tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_-infty^infty (-xe^-x^2)sin(tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(tx)$ and $v=frac12e^-x^2$ we obtain the ODE
$$f'(t)=-tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^-t^2tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_-infty^infty e^-(x-it)^2,dx=sqrt pi$$
answered Jul 21 at 18:05
Mark Viola
126k1172167
edited Jul 21 at 18:10
answered Jul 21 at 18:05
Mark Viola
126k1172167
answered Jul 21 at 18:05
Mark Viola
126k1172167
answered Jul 21 at 18:05
Mark Viola
126k1172167
126k1172167
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
add a comment |Â
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
1
1
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 21 at 18:14
add a comment |Â
up vote
0
down vote
Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.
Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $operatornameIm(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $displaystyleoint_Gamma(R) e^-z^2, dz$, where $Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^-z^2$ along the vertical edges can be shown to be $O(e^-R^2)$ as $R to infty$. Since $e^-z^2$ is entire, Cauchy's theorem gives $displaystyleoint_Gamma(R) e^-z^2, dz = 0$. Hence $displaystylelim_Rto infty int_-R - it^R - it e^-z^2, dz = lim_Rto infty int_-R^R e^-x^2, dx$, or $$int_-infty^infty e^-(x-it)^2, dx = int_-infty^infty e^-x^2, dx$$A similar argument holds when $t < 0$.
answered Jul 22 at 20:45
kobe
33.9k22146
add a comment |Â
up vote
0
down vote
Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.
Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $operatornameIm(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $displaystyleoint_Gamma(R) e^-z^2, dz$, where $Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^-z^2$ along the vertical edges can be shown to be $O(e^-R^2)$ as $R to infty$. Since $e^-z^2$ is entire, Cauchy's theorem gives $displaystyleoint_Gamma(R) e^-z^2, dz = 0$. Hence $displaystylelim_Rto infty int_-R - it^R - it e^-z^2, dz = lim_Rto infty int_-R^R e^-x^2, dx$, or $$int_-infty^infty e^-(x-it)^2, dx = int_-infty^infty e^-x^2, dx$$A similar argument holds when $t < 0$.
answered Jul 22 at 20:45
kobe
33.9k22146
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.
Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $operatornameIm(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $displaystyleoint_Gamma(R) e^-z^2, dz$, where $Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^-z^2$ along the vertical edges can be shown to be $O(e^-R^2)$ as $R to infty$. Since $e^-z^2$ is entire, Cauchy's theorem gives $displaystyleoint_Gamma(R) e^-z^2, dz = 0$. Hence $displaystylelim_Rto infty int_-R - it^R - it e^-z^2, dz = lim_Rto infty int_-R^R e^-x^2, dx$, or $$int_-infty^infty e^-(x-it)^2, dx = int_-infty^infty e^-x^2, dx$$A similar argument holds when $t < 0$.
answered Jul 22 at 20:45
kobe
33.9k22146
Another short cut is to use the substitution $y = x-it$, which converts the integral into a standard integral on the real line.
Unless $t = 0$, the substitution transforms the integral into an integral over the horizontal line $operatornameIm(z) = -t$, not over the real line. This method is really part of the contour integral method. If $t > 0$, we consider the contour integral $displaystyleoint_Gamma(R) e^-z^2, dz$, where $Gamma(R)$ is a clockwise-oriented rectangle in the complex plane with vertices at $-R, R, R - it$, and $-R - it$. The integrals of $e^-z^2$ along the vertical edges can be shown to be $O(e^-R^2)$ as $R to infty$. Since $e^-z^2$ is entire, Cauchy's theorem gives $displaystyleoint_Gamma(R) e^-z^2, dz = 0$. Hence $displaystylelim_Rto infty int_-R - it^R - it e^-z^2, dz = lim_Rto infty int_-R^R e^-x^2, dx$, or $$int_-infty^infty e^-(x-it)^2, dx = int_-infty^infty e^-x^2, dx$$A similar argument holds when $t < 0$.
answered Jul 22 at 20:45
kobe
33.9k22146
answered Jul 22 at 20:45
kobe
33.9k22146
answered Jul 22 at 20:45
kobe
33.9k22146
answered Jul 22 at 20:45
kobe
33.9k22146
33.9k22146
add a comment |Â
add a comment |Â
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The substitution doesn't convert the integral into one on the real line as the integration path is a line parallel to the real axis.
– mathcounterexamples.net
Jul 21 at 16:03
The substitution seems to shift the parallel line to the real line. The final integral is $int e^-y^2 dy$.
– JohnKnoxV
Jul 21 at 16:05
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
– mathcounterexamples.net
Jul 21 at 16:08