Mixing
ML inequality on $|int_{gamma} e^{iaz^{2}}dz|$ such that $z=Re^{itheta},theta in [0,frac{pi}{4}]$
$begingroup$
On a contour defined by sector of radius $R$ making an angle of $frac{pi}{4}$, I applied the ML inequality on the curvy(angular) part ofcontour and this is what i found $|int_{gamma} e^{iaz^{2}}dz|leqfrac{pi R}{4}$ since $|f(z)|leq 1$ and $L=frac{pi R}{4}$
But this doesn't seem right as $lim_{Rtoinfty}frac{pi R}{4}=infty$,even though I know that it should go to zero.
I am evaluating the fresnel integrals using this.
complex-analysis contour-integration
edited Jan 28 at 13:25
Dylan
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
$endgroup$
add a comment |
$begingroup$
On a contour defined by sector of radius $R$ making an angle of $frac{pi}{4}$, I applied the ML inequality on the curvy(angular) part ofcontour and this is what i found $|int_{gamma} e^{iaz^{2}}dz|leqfrac{pi R}{4}$ since $|f(z)|leq 1$ and $L=frac{pi R}{4}$
But this doesn't seem right as $lim_{Rtoinfty}frac{pi R}{4}=infty$,even though I know that it should go to zero.
I am evaluating the fresnel integrals using this.
complex-analysis contour-integration
edited Jan 28 at 13:25
Dylan
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
$endgroup$
add a comment |
$begingroup$
On a contour defined by sector of radius $R$ making an angle of $frac{pi}{4}$, I applied the ML inequality on the curvy(angular) part ofcontour and this is what i found $|int_{gamma} e^{iaz^{2}}dz|leqfrac{pi R}{4}$ since $|f(z)|leq 1$ and $L=frac{pi R}{4}$
But this doesn't seem right as $lim_{Rtoinfty}frac{pi R}{4}=infty$,even though I know that it should go to zero.
I am evaluating the fresnel integrals using this.
complex-analysis contour-integration
edited Jan 28 at 13:25
Dylan
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
$endgroup$
On a contour defined by sector of radius $R$ making an angle of $frac{pi}{4}$, I applied the ML inequality on the curvy(angular) part ofcontour and this is what i found $|int_{gamma} e^{iaz^{2}}dz|leqfrac{pi R}{4}$ since $|f(z)|leq 1$ and $L=frac{pi R}{4}$
But this doesn't seem right as $lim_{Rtoinfty}frac{pi R}{4}=infty$,even though I know that it should go to zero.
I am evaluating the fresnel integrals using this.
complex-analysis contour-integration
complex-analysis contour-integration
edited Jan 28 at 13:25
Dylan
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
edited Jan 28 at 13:25
Dylan
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
edited Jan 28 at 13:25
Dylan
14.2k31127
edited Jan 28 at 13:25
Dylan
14.2k31127
edited Jan 28 at 13:25
Dylan
14.2k31127
14.2k31127
asked Jan 28 at 9:29
ben tenysonben tenyson
414
asked Jan 28 at 9:29
ben tenysonben tenyson
414
asked Jan 28 at 9:29
ben tenysonben tenyson
414
414
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Note that
$$ leftvert int_{gamma} e^{iaz^2} dz rightvert le int_{gamma} leftvert e^{iaz^2} rightvert leftvert dz rightvert = int_0^{pi/4} e^{-aR^2sin 2theta} R dtheta $$
You can show geometrically that on $theta in [0,pi/4]$, the function $g(theta) = sin 2theta$ lies completely above its secant line, i.e.
$$ sin 2theta ge frac{4}{pi}theta implies e^{-aR^2sin2theta} le e^{-4aR^2 theta /pi} $$
Hence, the integral is bounded above by
$$ int_0^{pi/4} e^{-4aR^2 theta /pi} R dtheta = frac{pi}{4aR}left(1 - e^{-aR^2}right) $$
which tends to $0$ as $R to infty$
answered Jan 28 at 13:33
DylanDylan
14.2k31127
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Note that
$$ leftvert int_{gamma} e^{iaz^2} dz rightvert le int_{gamma} leftvert e^{iaz^2} rightvert leftvert dz rightvert = int_0^{pi/4} e^{-aR^2sin 2theta} R dtheta $$
You can show geometrically that on $theta in [0,pi/4]$, the function $g(theta) = sin 2theta$ lies completely above its secant line, i.e.
$$ sin 2theta ge frac{4}{pi}theta implies e^{-aR^2sin2theta} le e^{-4aR^2 theta /pi} $$
Hence, the integral is bounded above by
$$ int_0^{pi/4} e^{-4aR^2 theta /pi} R dtheta = frac{pi}{4aR}left(1 - e^{-aR^2}right) $$
which tends to $0$ as $R to infty$
answered Jan 28 at 13:33
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Note that
$$ leftvert int_{gamma} e^{iaz^2} dz rightvert le int_{gamma} leftvert e^{iaz^2} rightvert leftvert dz rightvert = int_0^{pi/4} e^{-aR^2sin 2theta} R dtheta $$
You can show geometrically that on $theta in [0,pi/4]$, the function $g(theta) = sin 2theta$ lies completely above its secant line, i.e.
$$ sin 2theta ge frac{4}{pi}theta implies e^{-aR^2sin2theta} le e^{-4aR^2 theta /pi} $$
Hence, the integral is bounded above by
$$ int_0^{pi/4} e^{-4aR^2 theta /pi} R dtheta = frac{pi}{4aR}left(1 - e^{-aR^2}right) $$
which tends to $0$ as $R to infty$
answered Jan 28 at 13:33
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Note that
$$ leftvert int_{gamma} e^{iaz^2} dz rightvert le int_{gamma} leftvert e^{iaz^2} rightvert leftvert dz rightvert = int_0^{pi/4} e^{-aR^2sin 2theta} R dtheta $$
You can show geometrically that on $theta in [0,pi/4]$, the function $g(theta) = sin 2theta$ lies completely above its secant line, i.e.
$$ sin 2theta ge frac{4}{pi}theta implies e^{-aR^2sin2theta} le e^{-4aR^2 theta /pi} $$
Hence, the integral is bounded above by
$$ int_0^{pi/4} e^{-4aR^2 theta /pi} R dtheta = frac{pi}{4aR}left(1 - e^{-aR^2}right) $$
which tends to $0$ as $R to infty$
answered Jan 28 at 13:33
DylanDylan
14.2k31127
$endgroup$
Note that
$$ leftvert int_{gamma} e^{iaz^2} dz rightvert le int_{gamma} leftvert e^{iaz^2} rightvert leftvert dz rightvert = int_0^{pi/4} e^{-aR^2sin 2theta} R dtheta $$
You can show geometrically that on $theta in [0,pi/4]$, the function $g(theta) = sin 2theta$ lies completely above its secant line, i.e.
$$ sin 2theta ge frac{4}{pi}theta implies e^{-aR^2sin2theta} le e^{-4aR^2 theta /pi} $$
Hence, the integral is bounded above by
$$ int_0^{pi/4} e^{-4aR^2 theta /pi} R dtheta = frac{pi}{4aR}left(1 - e^{-aR^2}right) $$
which tends to $0$ as $R to infty$
answered Jan 28 at 13:33
DylanDylan
14.2k31127
answered Jan 28 at 13:33
DylanDylan
14.2k31127
answered Jan 28 at 13:33
DylanDylan
14.2k31127
answered Jan 28 at 13:33
DylanDylan
14.2k31127
14.2k31127
add a comment |
add a comment |
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