How to integrate Fresnel Integrals? $int_0^y e^frac{-jbeta(z)^2}rho dz$
$begingroup$
I am having trouble solving this integration of a spherical fresnel zone with radius y
$displaystyleint_0^y e^frac{-jbeta(z)^2}rho dz$
, where j is complex and $beta$ and $rho$ are constants.
I have seen other questions such as
$int e^{ix^2}dx$
Complex Integration
but I do not understand how the bounds in that example were obtained when converted to polar coordinates. I have also seen other references that say to use euler's relationship but that leads to a very complicated result using the maclaurin series,
I know I have to use u-substitution to get
$int_0^y e^{-ju^2}du$
But don't know how to continue. Any help is appreciated!
calculus integration complex-analysis mathematical-physics electromagnetism
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
$endgroup$
add a comment |
$begingroup$
I am having trouble solving this integration of a spherical fresnel zone with radius y
$displaystyleint_0^y e^frac{-jbeta(z)^2}rho dz$
, where j is complex and $beta$ and $rho$ are constants.
I have seen other questions such as
$int e^{ix^2}dx$
Complex Integration
but I do not understand how the bounds in that example were obtained when converted to polar coordinates. I have also seen other references that say to use euler's relationship but that leads to a very complicated result using the maclaurin series,
I know I have to use u-substitution to get
$int_0^y e^{-ju^2}du$
But don't know how to continue. Any help is appreciated!
calculus integration complex-analysis mathematical-physics electromagnetism
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
$endgroup$
$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47
add a comment |
$begingroup$
I am having trouble solving this integration of a spherical fresnel zone with radius y
$displaystyleint_0^y e^frac{-jbeta(z)^2}rho dz$
, where j is complex and $beta$ and $rho$ are constants.
I have seen other questions such as
$int e^{ix^2}dx$
Complex Integration
but I do not understand how the bounds in that example were obtained when converted to polar coordinates. I have also seen other references that say to use euler's relationship but that leads to a very complicated result using the maclaurin series,
I know I have to use u-substitution to get
$int_0^y e^{-ju^2}du$
But don't know how to continue. Any help is appreciated!
calculus integration complex-analysis mathematical-physics electromagnetism
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
$endgroup$
I am having trouble solving this integration of a spherical fresnel zone with radius y
$displaystyleint_0^y e^frac{-jbeta(z)^2}rho dz$
, where j is complex and $beta$ and $rho$ are constants.
I have seen other questions such as
$int e^{ix^2}dx$
Complex Integration
but I do not understand how the bounds in that example were obtained when converted to polar coordinates. I have also seen other references that say to use euler's relationship but that leads to a very complicated result using the maclaurin series,
I know I have to use u-substitution to get
$int_0^y e^{-ju^2}du$
But don't know how to continue. Any help is appreciated!
calculus integration complex-analysis mathematical-physics electromagnetism
calculus integration complex-analysis mathematical-physics electromagnetism
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
edited Dec 5 '18 at 7:40
Dylan
13.4k31027
13.4k31027
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
asked Dec 4 '18 at 22:33
It'sJohnnyIt'sJohnny
12
12
$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47
add a comment |
$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47
$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47
add a comment |
1 Answer
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$begingroup$
$$int_0^yint_0^{2pi} e^{-jr^2}rdrdtheta$$
Is only equal to $I^2$ in the limit $y to infty$
Notice that the region of integration was a square and has now become a circle.
I'm fairly certain that $ int_0 ^y e^{-ax^2} dx$ cannot be expressed in terms of elementary functions, its solution is sometimes referred to as the error function.
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$$int_0^yint_0^{2pi} e^{-jr^2}rdrdtheta$$
Is only equal to $I^2$ in the limit $y to infty$
Notice that the region of integration was a square and has now become a circle.
I'm fairly certain that $ int_0 ^y e^{-ax^2} dx$ cannot be expressed in terms of elementary functions, its solution is sometimes referred to as the error function.
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
$endgroup$
add a comment |
$begingroup$
$$int_0^yint_0^{2pi} e^{-jr^2}rdrdtheta$$
Is only equal to $I^2$ in the limit $y to infty$
Notice that the region of integration was a square and has now become a circle.
I'm fairly certain that $ int_0 ^y e^{-ax^2} dx$ cannot be expressed in terms of elementary functions, its solution is sometimes referred to as the error function.
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
$endgroup$
add a comment |
$begingroup$
$$int_0^yint_0^{2pi} e^{-jr^2}rdrdtheta$$
Is only equal to $I^2$ in the limit $y to infty$
Notice that the region of integration was a square and has now become a circle.
I'm fairly certain that $ int_0 ^y e^{-ax^2} dx$ cannot be expressed in terms of elementary functions, its solution is sometimes referred to as the error function.
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
$endgroup$
$$int_0^yint_0^{2pi} e^{-jr^2}rdrdtheta$$
Is only equal to $I^2$ in the limit $y to infty$
Notice that the region of integration was a square and has now become a circle.
I'm fairly certain that $ int_0 ^y e^{-ax^2} dx$ cannot be expressed in terms of elementary functions, its solution is sometimes referred to as the error function.
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
answered Dec 4 '18 at 22:52
WW1WW1
7,3251712
7,3251712
add a comment |
add a comment |
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$begingroup$
There is at least a problem with calculation of the last integral: $j$ disapeared
$endgroup$
– Damien
Dec 4 '18 at 22:44
$begingroup$
Another problem: when integrating, the square becomes a circle ...
$endgroup$
– Damien
Dec 4 '18 at 22:47