Mixing
Contour Integrantion of a exponential function
$begingroup$
I am trying to evaluate an integral of type
$$ I = int_{-infty}^{infty} frac{e^{ikx}P(x)}{Q(x)} ,dx$$
where
$P(x), Q(x)$ are polynomials;
$Q(x)$ has no zeros on the real line;- $mathoperator{Order}(P(x)) + 2 > mathoperator{Order}(Q(x)).$
Now according to theory, If I want the contribution of the curved portion of the contour to be zero I have to choose a semi-circular contour lying on top of the Real axis if $k > 0,$ and choose a semi-circular contour lying below the Real axis if $k < 0.$ These contours are shown below:
Fig 1: Upper Semi-Circular Contour ( for k > 0 )
Fig 2: Lower semicircular contour ( for k < 0 )
Now to prove that the curved section of the contour does not contribute I need to compute an ML estimate.Now let c = $left|kright|$.Regardless of which countour we pick in each of the cases:
Case 1 ( k > 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Case 2 ( k < 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{-icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Where $z = Re^{itheta}$ in both cases.Which seems to imply that the choice of contour does not matter because $|e^{icx}|$ =
$|e^{-icx}|$ = 1, so the two ML estimates will always go to the same value.How is this reasoning wrong? Is there a special reason why the ML estimate over the curved portion will only go to zero if I pick the appropriate semi-circular contour?
integration complex-analysis complex-integration
edited Jan 12 at 14:15
amWhy
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate an integral of type
$$ I = int_{-infty}^{infty} frac{e^{ikx}P(x)}{Q(x)} ,dx$$
where
$P(x), Q(x)$ are polynomials;
$Q(x)$ has no zeros on the real line;- $mathoperator{Order}(P(x)) + 2 > mathoperator{Order}(Q(x)).$
Now according to theory, If I want the contribution of the curved portion of the contour to be zero I have to choose a semi-circular contour lying on top of the Real axis if $k > 0,$ and choose a semi-circular contour lying below the Real axis if $k < 0.$ These contours are shown below:
Fig 1: Upper Semi-Circular Contour ( for k > 0 )
Fig 2: Lower semicircular contour ( for k < 0 )
Now to prove that the curved section of the contour does not contribute I need to compute an ML estimate.Now let c = $left|kright|$.Regardless of which countour we pick in each of the cases:
Case 1 ( k > 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Case 2 ( k < 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{-icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Where $z = Re^{itheta}$ in both cases.Which seems to imply that the choice of contour does not matter because $|e^{icx}|$ =
$|e^{-icx}|$ = 1, so the two ML estimates will always go to the same value.How is this reasoning wrong? Is there a special reason why the ML estimate over the curved portion will only go to zero if I pick the appropriate semi-circular contour?
integration complex-analysis complex-integration
edited Jan 12 at 14:15
amWhy
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate an integral of type
$$ I = int_{-infty}^{infty} frac{e^{ikx}P(x)}{Q(x)} ,dx$$
where
$P(x), Q(x)$ are polynomials;
$Q(x)$ has no zeros on the real line;- $mathoperator{Order}(P(x)) + 2 > mathoperator{Order}(Q(x)).$
Now according to theory, If I want the contribution of the curved portion of the contour to be zero I have to choose a semi-circular contour lying on top of the Real axis if $k > 0,$ and choose a semi-circular contour lying below the Real axis if $k < 0.$ These contours are shown below:
Fig 1: Upper Semi-Circular Contour ( for k > 0 )
Fig 2: Lower semicircular contour ( for k < 0 )
Now to prove that the curved section of the contour does not contribute I need to compute an ML estimate.Now let c = $left|kright|$.Regardless of which countour we pick in each of the cases:
Case 1 ( k > 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Case 2 ( k < 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{-icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Where $z = Re^{itheta}$ in both cases.Which seems to imply that the choice of contour does not matter because $|e^{icx}|$ =
$|e^{-icx}|$ = 1, so the two ML estimates will always go to the same value.How is this reasoning wrong? Is there a special reason why the ML estimate over the curved portion will only go to zero if I pick the appropriate semi-circular contour?
integration complex-analysis complex-integration
edited Jan 12 at 14:15
amWhy
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
$endgroup$
I am trying to evaluate an integral of type
$$ I = int_{-infty}^{infty} frac{e^{ikx}P(x)}{Q(x)} ,dx$$
where
$P(x), Q(x)$ are polynomials;
$Q(x)$ has no zeros on the real line;- $mathoperator{Order}(P(x)) + 2 > mathoperator{Order}(Q(x)).$
Now according to theory, If I want the contribution of the curved portion of the contour to be zero I have to choose a semi-circular contour lying on top of the Real axis if $k > 0,$ and choose a semi-circular contour lying below the Real axis if $k < 0.$ These contours are shown below:
Fig 1: Upper Semi-Circular Contour ( for k > 0 )
Fig 2: Lower semicircular contour ( for k < 0 )
Now to prove that the curved section of the contour does not contribute I need to compute an ML estimate.Now let c = $left|kright|$.Regardless of which countour we pick in each of the cases:
Case 1 ( k > 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Case 2 ( k < 0 ) :
$$
left| int_{gamma _R} f(z) dz right| leq ML
$$
$$
left| int_{gamma _R} f(z) dz right| leq
lim_{R - > infty} left [
frac{left|e^{-icz}right|
left| P(z) right|}
{left| Q(z) right|} right] * pi R
$$
Where $z = Re^{itheta}$ in both cases.Which seems to imply that the choice of contour does not matter because $|e^{icx}|$ =
$|e^{-icx}|$ = 1, so the two ML estimates will always go to the same value.How is this reasoning wrong? Is there a special reason why the ML estimate over the curved portion will only go to zero if I pick the appropriate semi-circular contour?
integration complex-analysis complex-integration
integration complex-analysis complex-integration
edited Jan 12 at 14:15
amWhy
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
edited Jan 12 at 14:15
amWhy
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
edited Jan 12 at 14:15
amWhy
1
edited Jan 12 at 14:15
amWhy
1
edited Jan 12 at 14:15
amWhy
1
1
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
asked Mar 16 '14 at 17:19
KillaKemKillaKem
2751416
2751416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming $k$ is real : $|exp(ikz)|=|exp(ik(x+iy))|=exp(-ky)$. This expression is less than $1$ if $k>0$ and $y>0$ or if $k<0$ and $y<0$. Thus you want to use a semicircle in the upper halfplane when $k>0$ and a semicricle in the lower halfplane when $k<0$.
If $k$ is not real, your integral won't converge.
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
$endgroup$
add a comment |
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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
$begingroup$
Assuming $k$ is real : $|exp(ikz)|=|exp(ik(x+iy))|=exp(-ky)$. This expression is less than $1$ if $k>0$ and $y>0$ or if $k<0$ and $y<0$. Thus you want to use a semicircle in the upper halfplane when $k>0$ and a semicricle in the lower halfplane when $k<0$.
If $k$ is not real, your integral won't converge.
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
Assuming $k$ is real : $|exp(ikz)|=|exp(ik(x+iy))|=exp(-ky)$. This expression is less than $1$ if $k>0$ and $y>0$ or if $k<0$ and $y<0$. Thus you want to use a semicircle in the upper halfplane when $k>0$ and a semicricle in the lower halfplane when $k<0$.
If $k$ is not real, your integral won't converge.
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
$endgroup$
add a comment |
$begingroup$
Assuming $k$ is real : $|exp(ikz)|=|exp(ik(x+iy))|=exp(-ky)$. This expression is less than $1$ if $k>0$ and $y>0$ or if $k<0$ and $y<0$. Thus you want to use a semicircle in the upper halfplane when $k>0$ and a semicricle in the lower halfplane when $k<0$.
If $k$ is not real, your integral won't converge.
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
$endgroup$
Assuming $k$ is real : $|exp(ikz)|=|exp(ik(x+iy))|=exp(-ky)$. This expression is less than $1$ if $k>0$ and $y>0$ or if $k<0$ and $y<0$. Thus you want to use a semicircle in the upper halfplane when $k>0$ and a semicricle in the lower halfplane when $k<0$.
If $k$ is not real, your integral won't converge.
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
answered Mar 16 '14 at 18:24
mrfmrf
37.4k54685
37.4k54685
add a comment |
add a comment |
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