Mixing
Characteristic Function of Gaussian/ Change of Variables in Complex Plane
$begingroup$
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 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
$endgroup$
$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08
add a comment |
$begingroup$
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 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
$endgroup$
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
probability complex-analysis analysis
asked Jul 21 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
asked Jul 21 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
asked Jul 21 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
asked Jul 21 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
asked Jul 21 '18 at 15:46
JohnKnoxVJohnKnoxV
517213
517213
$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08
add a comment |
$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08
$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is one approach that does not require a change of variables and contour deformation. First we write
$$begin{align}
int_{-infty}^infty e^{-(x-it)^2},dx&=e^{t^2}int_{-infty}^infty e^{-x^2+i2xt},dx\\
&=e^{t^2}int_{-infty}^infty e^{-x^2}cos(2tx),dxtag1
end{align}$$
Let $f(t)$ be given by the integral
$$f(t)=int_{-infty}^infty e^{-x^2}cos(2tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_{-infty}^infty (-2xe^{-x^2})sin(2tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE
$$f'(t)=-2tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^{-t^2}tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_{-infty}^infty e^{-(x-it)^2},dx=sqrt pi$$
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
$endgroup$
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
add a comment |
$begingroup$
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 $operatorname{Im}(z) = -t$, 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 '18 at 20:45
kobekobe
35.1k22248
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is one approach that does not require a change of variables and contour deformation. First we write
$$begin{align}
int_{-infty}^infty e^{-(x-it)^2},dx&=e^{t^2}int_{-infty}^infty e^{-x^2+i2xt},dx\\
&=e^{t^2}int_{-infty}^infty e^{-x^2}cos(2tx),dxtag1
end{align}$$
Let $f(t)$ be given by the integral
$$f(t)=int_{-infty}^infty e^{-x^2}cos(2tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_{-infty}^infty (-2xe^{-x^2})sin(2tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE
$$f'(t)=-2tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^{-t^2}tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_{-infty}^infty e^{-(x-it)^2},dx=sqrt pi$$
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
$endgroup$
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
add a comment |
$begingroup$
Here is one approach that does not require a change of variables and contour deformation. First we write
$$begin{align}
int_{-infty}^infty e^{-(x-it)^2},dx&=e^{t^2}int_{-infty}^infty e^{-x^2+i2xt},dx\\
&=e^{t^2}int_{-infty}^infty e^{-x^2}cos(2tx),dxtag1
end{align}$$
Let $f(t)$ be given by the integral
$$f(t)=int_{-infty}^infty e^{-x^2}cos(2tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_{-infty}^infty (-2xe^{-x^2})sin(2tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE
$$f'(t)=-2tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^{-t^2}tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_{-infty}^infty e^{-(x-it)^2},dx=sqrt pi$$
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
$endgroup$
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
add a comment |
$begingroup$
Here is one approach that does not require a change of variables and contour deformation. First we write
$$begin{align}
int_{-infty}^infty e^{-(x-it)^2},dx&=e^{t^2}int_{-infty}^infty e^{-x^2+i2xt},dx\\
&=e^{t^2}int_{-infty}^infty e^{-x^2}cos(2tx),dxtag1
end{align}$$
Let $f(t)$ be given by the integral
$$f(t)=int_{-infty}^infty e^{-x^2}cos(2tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_{-infty}^infty (-2xe^{-x^2})sin(2tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE
$$f'(t)=-2tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^{-t^2}tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_{-infty}^infty e^{-(x-it)^2},dx=sqrt pi$$
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
$endgroup$
Here is one approach that does not require a change of variables and contour deformation. First we write
$$begin{align}
int_{-infty}^infty e^{-(x-it)^2},dx&=e^{t^2}int_{-infty}^infty e^{-x^2+i2xt},dx\\
&=e^{t^2}int_{-infty}^infty e^{-x^2}cos(2tx),dxtag1
end{align}$$
Let $f(t)$ be given by the integral
$$f(t)=int_{-infty}^infty e^{-x^2}cos(2tx),dxtag2$$
Differentiation reveals of $(2)$
$$f'(t)=int_{-infty}^infty (-2xe^{-x^2})sin(2tx),dxtag3$$
Integrating by parts the integral in $(3)$ with $u=sin(2tx)$ and $v=e^{-x^2}$ we obtain the ODE
$$f'(t)=-2tf(t) tag4$$
Using $f(0)=sqrt pi$, we find from $(4)$ that
$$f(t)=sqrtpi e^{-t^2}tag5$$
Substituting $(5)$ into $(1)$ yields the coveted result
$$int_{-infty}^infty e^{-(x-it)^2},dx=sqrt pi$$
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
edited Feb 2 at 21:11
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
answered Jul 21 '18 at 18:05
Mark ViolaMark Viola
134k1278177
134k1278177
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
add a comment |
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
1
1
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Jul 21 '18 at 18:14
add a comment |
$begingroup$
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 $operatorname{Im}(z) = -t$, 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 '18 at 20:45
kobekobe
35.1k22248
$endgroup$
add a comment |
$begingroup$
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 $operatorname{Im}(z) = -t$, 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 '18 at 20:45
kobekobe
35.1k22248
$endgroup$
add a comment |
$begingroup$
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 $operatorname{Im}(z) = -t$, 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 '18 at 20:45
kobekobe
35.1k22248
$endgroup$
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 $operatorname{Im}(z) = -t$, 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 '18 at 20:45
kobekobe
35.1k22248
answered Jul 22 '18 at 20:45
kobekobe
35.1k22248
answered Jul 22 '18 at 20:45
kobekobe
35.1k22248
answered Jul 22 '18 at 20:45
kobekobe
35.1k22248
35.1k22248
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:03
$begingroup$
The substitution seems to shift the parallel line to the real line. The final integral is $int e^{-y^2} dy$.
$endgroup$
– JohnKnoxV
Jul 21 '18 at 16:05
$begingroup$
No because the integration path is not $mathbb R$ but the line $x-it$ for $x in mathbb R$.
$endgroup$
– mathcounterexamples.net
Jul 21 '18 at 16:08