Uniform asymptotic expansion of integral with corner singularitiesAsymptotic expansion of $x_n$, $x_n=frac1tan(x_n)$Integral asymptotic expansion of $int_0^pi/2 exp(-xt^3cos t)dt$ as $x to infty$The Asymptotic Expansion of The Exponential IntegralAsymptotic Expansion of an Integral involving Modified Bessel FunctionsAsymptotic expansion of a Laplace-type integral with a “manifold of maxima”Asymptotic expansion of Polygamma functionsUniform Convergence of a Asymptotic Series (Asymptotic Expansion of Integrals)Using asymptotic expansion of integralLaplace method (or other integral asymptotic) with near-cornerAsymptotic expansion of Erfi(x)
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Uniform asymptotic expansion of integral with corner singularities
Asymptotic expansion of $x_n$, $x_n=frac1tan(x_n)$Integral asymptotic expansion of $int_0^pi/2 exp(-xt^3cos t)dt$ as $x to infty$The Asymptotic Expansion of The Exponential IntegralAsymptotic Expansion of an Integral involving Modified Bessel FunctionsAsymptotic expansion of a Laplace-type integral with a “manifold of maxima”Asymptotic expansion of Polygamma functionsUniform Convergence of a Asymptotic Series (Asymptotic Expansion of Integrals)Using asymptotic expansion of integralLaplace method (or other integral asymptotic) with near-cornerAsymptotic expansion of Erfi(x)
$begingroup$
Let $f : mathbbR to mathbbR$ be a given function which is $1$-periodic and is smooth except for finitely many corner singularities in each period. An example is $-|sin(2pi x)|$, which has a corner singularity at $x=k/2$ for each integer $k$.
I'm now looking at
$$I(x,M)=int_-infty^infty e^-M f(x-y) dy$$
where $M$ is a large parameter. I would like to obtain an asymptotic expansion for $I$ as $M to infty$ which is uniform in $x$ when $f$ is fixed.
The difficulty in doing so is that approximating $f$ on an interval of length $O(M^-1)$ near $x$ cannot be done by a polynomial when $x$ is within $O(M^-1)$ of a singularity of $f$. This causes the Laplace method to fail to provide a uniform expansion, essentially because it provides two different expansions when $x$ is a singularity or not, and these do not agree with one another as $x$ passes through a singularity.
Generally when I see methods for uniform asymptotic expansions of integrals in the literature, they still rely on local analyticity assumptions in order to convert the problem into a complex analysis problem. Or they simply identify the problem with some special function and then perhaps perform further asymptotics directly on the special function. In any case, I don't see how this can work here. Is there some alternative method?
An alternative that comes to mind is to use integration by parts. Putting aside remainder estimation, a first step of integration by parts for the $y in [0,infty)$ integral can be done as follows. Identify $delta>0$ where the first singularity is located in $y$, then split the integral there. Now the first step of integration by parts gives only the "local" term $fracf(x)M$ which is exactly what the first step of the Laplace method would give. The next step gives more interesting terms:
$$I=fracf(x)M - frac1M int_0^infty e^-My f'(x-y) dy$$
Now when we split the second integral and integrate by parts, the boundary terms provide a term corresponding to $f'(x)$ from the left endpoint of integration and additionally provide a non-cancelling pair of terms from the jump in $f'$ at $y=delta$.
Can we continue, or is there some breakdown further along in the procedure? It seems to work, providing a term corresponding to the jump (if present) in each derivative at $y=delta$. And the expansion itself appears to be a nice function of $x$ as long as singularity-to-singularity is a full period (as in the example, which is actually best characterized as being $1/2$-periodic). If singularity-to-singularity is not a full period then we need to retain more terms in order to maintain continuity at the midpoint between two singularities, but that's a technical detail.
real-analysis asymptotics laplace-method
asked Apr 1 at 2:04
IanIan
68.9k25392
$endgroup$
add a comment |
$begingroup$
Let $f : mathbbR to mathbbR$ be a given function which is $1$-periodic and is smooth except for finitely many corner singularities in each period. An example is $-|sin(2pi x)|$, which has a corner singularity at $x=k/2$ for each integer $k$.
I'm now looking at
$$I(x,M)=int_-infty^infty e^-M f(x-y) dy$$
where $M$ is a large parameter. I would like to obtain an asymptotic expansion for $I$ as $M to infty$ which is uniform in $x$ when $f$ is fixed.
The difficulty in doing so is that approximating $f$ on an interval of length $O(M^-1)$ near $x$ cannot be done by a polynomial when $x$ is within $O(M^-1)$ of a singularity of $f$. This causes the Laplace method to fail to provide a uniform expansion, essentially because it provides two different expansions when $x$ is a singularity or not, and these do not agree with one another as $x$ passes through a singularity.
Generally when I see methods for uniform asymptotic expansions of integrals in the literature, they still rely on local analyticity assumptions in order to convert the problem into a complex analysis problem. Or they simply identify the problem with some special function and then perhaps perform further asymptotics directly on the special function. In any case, I don't see how this can work here. Is there some alternative method?
An alternative that comes to mind is to use integration by parts. Putting aside remainder estimation, a first step of integration by parts for the $y in [0,infty)$ integral can be done as follows. Identify $delta>0$ where the first singularity is located in $y$, then split the integral there. Now the first step of integration by parts gives only the "local" term $fracf(x)M$ which is exactly what the first step of the Laplace method would give. The next step gives more interesting terms:
$$I=fracf(x)M - frac1M int_0^infty e^-My f'(x-y) dy$$
Now when we split the second integral and integrate by parts, the boundary terms provide a term corresponding to $f'(x)$ from the left endpoint of integration and additionally provide a non-cancelling pair of terms from the jump in $f'$ at $y=delta$.
Can we continue, or is there some breakdown further along in the procedure? It seems to work, providing a term corresponding to the jump (if present) in each derivative at $y=delta$. And the expansion itself appears to be a nice function of $x$ as long as singularity-to-singularity is a full period (as in the example, which is actually best characterized as being $1/2$-periodic). If singularity-to-singularity is not a full period then we need to retain more terms in order to maintain continuity at the midpoint between two singularities, but that's a technical detail.
real-analysis asymptotics laplace-method
asked Apr 1 at 2:04
IanIan
68.9k25392
$endgroup$
$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20
add a comment |
$begingroup$
Let $f : mathbbR to mathbbR$ be a given function which is $1$-periodic and is smooth except for finitely many corner singularities in each period. An example is $-|sin(2pi x)|$, which has a corner singularity at $x=k/2$ for each integer $k$.
I'm now looking at
$$I(x,M)=int_-infty^infty e^-M f(x-y) dy$$
where $M$ is a large parameter. I would like to obtain an asymptotic expansion for $I$ as $M to infty$ which is uniform in $x$ when $f$ is fixed.
The difficulty in doing so is that approximating $f$ on an interval of length $O(M^-1)$ near $x$ cannot be done by a polynomial when $x$ is within $O(M^-1)$ of a singularity of $f$. This causes the Laplace method to fail to provide a uniform expansion, essentially because it provides two different expansions when $x$ is a singularity or not, and these do not agree with one another as $x$ passes through a singularity.
Generally when I see methods for uniform asymptotic expansions of integrals in the literature, they still rely on local analyticity assumptions in order to convert the problem into a complex analysis problem. Or they simply identify the problem with some special function and then perhaps perform further asymptotics directly on the special function. In any case, I don't see how this can work here. Is there some alternative method?
An alternative that comes to mind is to use integration by parts. Putting aside remainder estimation, a first step of integration by parts for the $y in [0,infty)$ integral can be done as follows. Identify $delta>0$ where the first singularity is located in $y$, then split the integral there. Now the first step of integration by parts gives only the "local" term $fracf(x)M$ which is exactly what the first step of the Laplace method would give. The next step gives more interesting terms:
$$I=fracf(x)M - frac1M int_0^infty e^-My f'(x-y) dy$$
Now when we split the second integral and integrate by parts, the boundary terms provide a term corresponding to $f'(x)$ from the left endpoint of integration and additionally provide a non-cancelling pair of terms from the jump in $f'$ at $y=delta$.
Can we continue, or is there some breakdown further along in the procedure? It seems to work, providing a term corresponding to the jump (if present) in each derivative at $y=delta$. And the expansion itself appears to be a nice function of $x$ as long as singularity-to-singularity is a full period (as in the example, which is actually best characterized as being $1/2$-periodic). If singularity-to-singularity is not a full period then we need to retain more terms in order to maintain continuity at the midpoint between two singularities, but that's a technical detail.
real-analysis asymptotics laplace-method
asked Apr 1 at 2:04
IanIan
68.9k25392
$endgroup$
Let $f : mathbbR to mathbbR$ be a given function which is $1$-periodic and is smooth except for finitely many corner singularities in each period. An example is $-|sin(2pi x)|$, which has a corner singularity at $x=k/2$ for each integer $k$.
I'm now looking at
$$I(x,M)=int_-infty^infty e^-M f(x-y) dy$$
where $M$ is a large parameter. I would like to obtain an asymptotic expansion for $I$ as $M to infty$ which is uniform in $x$ when $f$ is fixed.
The difficulty in doing so is that approximating $f$ on an interval of length $O(M^-1)$ near $x$ cannot be done by a polynomial when $x$ is within $O(M^-1)$ of a singularity of $f$. This causes the Laplace method to fail to provide a uniform expansion, essentially because it provides two different expansions when $x$ is a singularity or not, and these do not agree with one another as $x$ passes through a singularity.
Generally when I see methods for uniform asymptotic expansions of integrals in the literature, they still rely on local analyticity assumptions in order to convert the problem into a complex analysis problem. Or they simply identify the problem with some special function and then perhaps perform further asymptotics directly on the special function. In any case, I don't see how this can work here. Is there some alternative method?
An alternative that comes to mind is to use integration by parts. Putting aside remainder estimation, a first step of integration by parts for the $y in [0,infty)$ integral can be done as follows. Identify $delta>0$ where the first singularity is located in $y$, then split the integral there. Now the first step of integration by parts gives only the "local" term $fracf(x)M$ which is exactly what the first step of the Laplace method would give. The next step gives more interesting terms:
$$I=fracf(x)M - frac1M int_0^infty e^-My f'(x-y) dy$$
Now when we split the second integral and integrate by parts, the boundary terms provide a term corresponding to $f'(x)$ from the left endpoint of integration and additionally provide a non-cancelling pair of terms from the jump in $f'$ at $y=delta$.
Can we continue, or is there some breakdown further along in the procedure? It seems to work, providing a term corresponding to the jump (if present) in each derivative at $y=delta$. And the expansion itself appears to be a nice function of $x$ as long as singularity-to-singularity is a full period (as in the example, which is actually best characterized as being $1/2$-periodic). If singularity-to-singularity is not a full period then we need to retain more terms in order to maintain continuity at the midpoint between two singularities, but that's a technical detail.
real-analysis asymptotics laplace-method
real-analysis asymptotics laplace-method
asked Apr 1 at 2:04
IanIan
68.9k25392
asked Apr 1 at 2:04
IanIan
68.9k25392
edited Apr 2 at 22:08
Ian
asked Apr 1 at 2:04
IanIan
68.9k25392
asked Apr 1 at 2:04
IanIan
68.9k25392
asked Apr 1 at 2:04
IanIan
68.9k25392
68.9k25392
$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20
add a comment |
$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20
$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20
add a comment |
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$begingroup$
Can you not re-write the integral as $int_0^infty e^-M y [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion?
$endgroup$
– Chip
Apr 3 at 1:49
$begingroup$
@Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$.
$endgroup$
– Ian
Apr 3 at 2:15
$begingroup$
can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $it etc$...
$endgroup$
– Chip
Apr 3 at 2:41
$begingroup$
If one writes starting with my comments above $f(x)$ as its back Fourier transform of $tilde f(omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 int_0^1 domega e^I omega x fracMM^2+omega^2 tildef(omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + cal O(1/M^3)$?
$endgroup$
– Chip
Apr 3 at 3:02
$begingroup$
@Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous.
$endgroup$
– Ian
Apr 3 at 3:20