Mixing
Laplace method (or other integral asymptotic) with near-corner
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
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Consider the integral
$$int_-infty^infty exp(-sqrth^2+M^2x^2) dx.$$
Here $h$ is a small positive parameter and $M$ is a large positive parameter. I would like to obtain a "reasonably uniform" asymptotic approximation for this integral in the limit of large $M$ and small $h$, specifically when $h$ goes to zero before $M$ goes to infinity.
The difficulty is that the leading order part of the Laplace method sees $sqrth^2+M^2 x^2$ as $h+fracM^22h x^2$, a quadratic function, but in fact this approximation is only any good where $|x| ll h/M$. By contrast there is a significant contribution to the integration over an interval of length on the order of $1/M$, which is much larger. Higher order Taylor approximations never see this because they just keep on assuming that $|x| ll h/M$ and thus proceed to divide by larger and larger powers of $h$.
An obvious alternative is to sacrifice accuracy in this $O(h/M)$ vicinity of $0$, for example by suppressing $h^2$ altogether, but this obviously does not achieve $o(h)$ accuracy, which is required for my application. Is there another workaround for this situation? Perhaps by "matching" the two approximations which are valid in different regimes?
real-analysis integration asymptotics
asked Jul 19 at 15:03
Ian
65.1k24681
add a comment |Â
up vote
7
down vote
favorite
Consider the integral
$$int_-infty^infty exp(-sqrth^2+M^2x^2) dx.$$
Here $h$ is a small positive parameter and $M$ is a large positive parameter. I would like to obtain a "reasonably uniform" asymptotic approximation for this integral in the limit of large $M$ and small $h$, specifically when $h$ goes to zero before $M$ goes to infinity.
The difficulty is that the leading order part of the Laplace method sees $sqrth^2+M^2 x^2$ as $h+fracM^22h x^2$, a quadratic function, but in fact this approximation is only any good where $|x| ll h/M$. By contrast there is a significant contribution to the integration over an interval of length on the order of $1/M$, which is much larger. Higher order Taylor approximations never see this because they just keep on assuming that $|x| ll h/M$ and thus proceed to divide by larger and larger powers of $h$.
An obvious alternative is to sacrifice accuracy in this $O(h/M)$ vicinity of $0$, for example by suppressing $h^2$ altogether, but this obviously does not achieve $o(h)$ accuracy, which is required for my application. Is there another workaround for this situation? Perhaps by "matching" the two approximations which are valid in different regimes?
real-analysis integration asymptotics
asked Jul 19 at 15:03
Ian
65.1k24681
You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Consider the integral
$$int_-infty^infty exp(-sqrth^2+M^2x^2) dx.$$
Here $h$ is a small positive parameter and $M$ is a large positive parameter. I would like to obtain a "reasonably uniform" asymptotic approximation for this integral in the limit of large $M$ and small $h$, specifically when $h$ goes to zero before $M$ goes to infinity.
The difficulty is that the leading order part of the Laplace method sees $sqrth^2+M^2 x^2$ as $h+fracM^22h x^2$, a quadratic function, but in fact this approximation is only any good where $|x| ll h/M$. By contrast there is a significant contribution to the integration over an interval of length on the order of $1/M$, which is much larger. Higher order Taylor approximations never see this because they just keep on assuming that $|x| ll h/M$ and thus proceed to divide by larger and larger powers of $h$.
An obvious alternative is to sacrifice accuracy in this $O(h/M)$ vicinity of $0$, for example by suppressing $h^2$ altogether, but this obviously does not achieve $o(h)$ accuracy, which is required for my application. Is there another workaround for this situation? Perhaps by "matching" the two approximations which are valid in different regimes?
real-analysis integration asymptotics
asked Jul 19 at 15:03
Ian
65.1k24681
Consider the integral
$$int_-infty^infty exp(-sqrth^2+M^2x^2) dx.$$
Here $h$ is a small positive parameter and $M$ is a large positive parameter. I would like to obtain a "reasonably uniform" asymptotic approximation for this integral in the limit of large $M$ and small $h$, specifically when $h$ goes to zero before $M$ goes to infinity.
The difficulty is that the leading order part of the Laplace method sees $sqrth^2+M^2 x^2$ as $h+fracM^22h x^2$, a quadratic function, but in fact this approximation is only any good where $|x| ll h/M$. By contrast there is a significant contribution to the integration over an interval of length on the order of $1/M$, which is much larger. Higher order Taylor approximations never see this because they just keep on assuming that $|x| ll h/M$ and thus proceed to divide by larger and larger powers of $h$.
An obvious alternative is to sacrifice accuracy in this $O(h/M)$ vicinity of $0$, for example by suppressing $h^2$ altogether, but this obviously does not achieve $o(h)$ accuracy, which is required for my application. Is there another workaround for this situation? Perhaps by "matching" the two approximations which are valid in different regimes?
real-analysis integration asymptotics
asked Jul 19 at 15:03
Ian
65.1k24681
edited Jul 19 at 15:33
asked Jul 19 at 15:03
Ian
65.1k24681
asked Jul 19 at 15:03
Ian
65.1k24681
asked Jul 19 at 15:03
Ian
65.1k24681
65.1k24681
You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06
add a comment |Â
You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06
You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
As noticed, a simple Laplace method cannot be used here, as 2 scales are involved. A uniform asymptotic expansion should be found. Alternatively, in this case, we can recognize a modified Bessel function. Indeed, changing $x=frachMsinh t$, the integral can be written as
beginalign
I&=int_-infty^infty exp(-sqrth^2+M^2x^2) ,dx\
&=frachMint_-infty^infty exp(-hcosh t)cosh t ,dt
endalign
which is proportional to an integral representation of a modified Bessel function (DLMF):
beginequation
K_nuleft(zright)=int_0^inftye^-zcosh tcoshleft(nu tright)%
mathrmdt
endequation
with $nu=1,z=h$,
beginequation
I=frac2hMK_1(h)
endequation
Using the series expansion of the Bessel function near $h=0$ (DLMF),
beginalign
I&sim frac2hMleft[ frac1h+frach2left(lnfrach2 +gamma-frac12right)+ldotsright]\
&simfrac2M+frach^2Mleft(lnfrach2 +gamma-frac12right)+ldots
endalign
EDIT:Another method using the Mellin transform technique
Changing $x=th/M$ and then $u=sqrt1+t^2-1$, the problem is equivalent to find the small $h$ behavior of
beginalign
I&=2frachMint_0^inftyexp(-hsqrt1+t^2),dt\
&=2frachMe^-hint_0^inftyfracu+1sqrtu(u+2)exp(-hu),du
endalign
We have thus to find a Laplace transforms with a small parameter. A classical method which uses the Mellin transform technique is given in (DLMF). Intermediate results are given below, with the help of a CAS. Defining
beginequation
H(u)=fracu+1sqrtu(u+2)
endequation he following behaviors hold:
beginarraylll
H(u)&sim 1+frac12u^2+Oleft( u^-3 right)& text for utoinfty\
&sim O(u^-1/2)& text for uto 0
endarray
and the Mellin transform is
beginequation
mathcalMleft[H(u) right](z)=pi^-1/22^z-1(z-1)Gamma(-z)Gammaleft( z-frac12 right)
endequation
For the function
beginequation
F(z)=-h^z-1Gamma(1-z)mathcalMleft[H(u) right](z)
endequation
the residues for $z=0,1,2,3$ are
beginalign
left. operatornameresF(z)right|_z=0&=frac1h\
left. operatornameresF(z)right|_z=1&=1\
left. operatornameresF(z)right|_z=2&=frach2left[ lnfrach2+gamma+frac12right]\
left. operatornameresF(z)right|_z=3&=frach^212left[6 lnfrach2+6gamma-1right]
endalign
With $e^-h= 1-h+h^2/2+O(h^3)$,
beginalign
I&sim frac2hM(1-h+frach^22)left[ frac1h+frach2left( lnfrach2+gamma+frac12right)+frach^212left(6 lnfrach2+6gamma-1right)right]\
&sim frac2M+frach^2Mleft( lnfrach2+gamma-frac12 right)
endalign
answered Jul 19 at 15:39
Paul Enta
3,3271925
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
As noticed, a simple Laplace method cannot be used here, as 2 scales are involved. A uniform asymptotic expansion should be found. Alternatively, in this case, we can recognize a modified Bessel function. Indeed, changing $x=frachMsinh t$, the integral can be written as
beginalign
I&=int_-infty^infty exp(-sqrth^2+M^2x^2) ,dx\
&=frachMint_-infty^infty exp(-hcosh t)cosh t ,dt
endalign
which is proportional to an integral representation of a modified Bessel function (DLMF):
beginequation
K_nuleft(zright)=int_0^inftye^-zcosh tcoshleft(nu tright)%
mathrmdt
endequation
with $nu=1,z=h$,
beginequation
I=frac2hMK_1(h)
endequation
Using the series expansion of the Bessel function near $h=0$ (DLMF),
beginalign
I&sim frac2hMleft[ frac1h+frach2left(lnfrach2 +gamma-frac12right)+ldotsright]\
&simfrac2M+frach^2Mleft(lnfrach2 +gamma-frac12right)+ldots
endalign
EDIT:Another method using the Mellin transform technique
Changing $x=th/M$ and then $u=sqrt1+t^2-1$, the problem is equivalent to find the small $h$ behavior of
beginalign
I&=2frachMint_0^inftyexp(-hsqrt1+t^2),dt\
&=2frachMe^-hint_0^inftyfracu+1sqrtu(u+2)exp(-hu),du
endalign
We have thus to find a Laplace transforms with a small parameter. A classical method which uses the Mellin transform technique is given in (DLMF). Intermediate results are given below, with the help of a CAS. Defining
beginequation
H(u)=fracu+1sqrtu(u+2)
endequation he following behaviors hold:
beginarraylll
H(u)&sim 1+frac12u^2+Oleft( u^-3 right)& text for utoinfty\
&sim O(u^-1/2)& text for uto 0
endarray
and the Mellin transform is
beginequation
mathcalMleft[H(u) right](z)=pi^-1/22^z-1(z-1)Gamma(-z)Gammaleft( z-frac12 right)
endequation
For the function
beginequation
F(z)=-h^z-1Gamma(1-z)mathcalMleft[H(u) right](z)
endequation
the residues for $z=0,1,2,3$ are
beginalign
left. operatornameresF(z)right|_z=0&=frac1h\
left. operatornameresF(z)right|_z=1&=1\
left. operatornameresF(z)right|_z=2&=frach2left[ lnfrach2+gamma+frac12right]\
left. operatornameresF(z)right|_z=3&=frach^212left[6 lnfrach2+6gamma-1right]
endalign
With $e^-h= 1-h+h^2/2+O(h^3)$,
beginalign
I&sim frac2hM(1-h+frach^22)left[ frac1h+frach2left( lnfrach2+gamma+frac12right)+frach^212left(6 lnfrach2+6gamma-1right)right]\
&sim frac2M+frach^2Mleft( lnfrach2+gamma-frac12 right)
endalign
answered Jul 19 at 15:39
Paul Enta
3,3271925
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
add a comment |Â
up vote
7
down vote
accepted
As noticed, a simple Laplace method cannot be used here, as 2 scales are involved. A uniform asymptotic expansion should be found. Alternatively, in this case, we can recognize a modified Bessel function. Indeed, changing $x=frachMsinh t$, the integral can be written as
beginalign
I&=int_-infty^infty exp(-sqrth^2+M^2x^2) ,dx\
&=frachMint_-infty^infty exp(-hcosh t)cosh t ,dt
endalign
which is proportional to an integral representation of a modified Bessel function (DLMF):
beginequation
K_nuleft(zright)=int_0^inftye^-zcosh tcoshleft(nu tright)%
mathrmdt
endequation
with $nu=1,z=h$,
beginequation
I=frac2hMK_1(h)
endequation
Using the series expansion of the Bessel function near $h=0$ (DLMF),
beginalign
I&sim frac2hMleft[ frac1h+frach2left(lnfrach2 +gamma-frac12right)+ldotsright]\
&simfrac2M+frach^2Mleft(lnfrach2 +gamma-frac12right)+ldots
endalign
EDIT:Another method using the Mellin transform technique
Changing $x=th/M$ and then $u=sqrt1+t^2-1$, the problem is equivalent to find the small $h$ behavior of
beginalign
I&=2frachMint_0^inftyexp(-hsqrt1+t^2),dt\
&=2frachMe^-hint_0^inftyfracu+1sqrtu(u+2)exp(-hu),du
endalign
We have thus to find a Laplace transforms with a small parameter. A classical method which uses the Mellin transform technique is given in (DLMF). Intermediate results are given below, with the help of a CAS. Defining
beginequation
H(u)=fracu+1sqrtu(u+2)
endequation he following behaviors hold:
beginarraylll
H(u)&sim 1+frac12u^2+Oleft( u^-3 right)& text for utoinfty\
&sim O(u^-1/2)& text for uto 0
endarray
and the Mellin transform is
beginequation
mathcalMleft[H(u) right](z)=pi^-1/22^z-1(z-1)Gamma(-z)Gammaleft( z-frac12 right)
endequation
For the function
beginequation
F(z)=-h^z-1Gamma(1-z)mathcalMleft[H(u) right](z)
endequation
the residues for $z=0,1,2,3$ are
beginalign
left. operatornameresF(z)right|_z=0&=frac1h\
left. operatornameresF(z)right|_z=1&=1\
left. operatornameresF(z)right|_z=2&=frach2left[ lnfrach2+gamma+frac12right]\
left. operatornameresF(z)right|_z=3&=frach^212left[6 lnfrach2+6gamma-1right]
endalign
With $e^-h= 1-h+h^2/2+O(h^3)$,
beginalign
I&sim frac2hM(1-h+frach^22)left[ frac1h+frach2left( lnfrach2+gamma+frac12right)+frach^212left(6 lnfrach2+6gamma-1right)right]\
&sim frac2M+frach^2Mleft( lnfrach2+gamma-frac12 right)
endalign
answered Jul 19 at 15:39
Paul Enta
3,3271925
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
As noticed, a simple Laplace method cannot be used here, as 2 scales are involved. A uniform asymptotic expansion should be found. Alternatively, in this case, we can recognize a modified Bessel function. Indeed, changing $x=frachMsinh t$, the integral can be written as
beginalign
I&=int_-infty^infty exp(-sqrth^2+M^2x^2) ,dx\
&=frachMint_-infty^infty exp(-hcosh t)cosh t ,dt
endalign
which is proportional to an integral representation of a modified Bessel function (DLMF):
beginequation
K_nuleft(zright)=int_0^inftye^-zcosh tcoshleft(nu tright)%
mathrmdt
endequation
with $nu=1,z=h$,
beginequation
I=frac2hMK_1(h)
endequation
Using the series expansion of the Bessel function near $h=0$ (DLMF),
beginalign
I&sim frac2hMleft[ frac1h+frach2left(lnfrach2 +gamma-frac12right)+ldotsright]\
&simfrac2M+frach^2Mleft(lnfrach2 +gamma-frac12right)+ldots
endalign
EDIT:Another method using the Mellin transform technique
Changing $x=th/M$ and then $u=sqrt1+t^2-1$, the problem is equivalent to find the small $h$ behavior of
beginalign
I&=2frachMint_0^inftyexp(-hsqrt1+t^2),dt\
&=2frachMe^-hint_0^inftyfracu+1sqrtu(u+2)exp(-hu),du
endalign
We have thus to find a Laplace transforms with a small parameter. A classical method which uses the Mellin transform technique is given in (DLMF). Intermediate results are given below, with the help of a CAS. Defining
beginequation
H(u)=fracu+1sqrtu(u+2)
endequation he following behaviors hold:
beginarraylll
H(u)&sim 1+frac12u^2+Oleft( u^-3 right)& text for utoinfty\
&sim O(u^-1/2)& text for uto 0
endarray
and the Mellin transform is
beginequation
mathcalMleft[H(u) right](z)=pi^-1/22^z-1(z-1)Gamma(-z)Gammaleft( z-frac12 right)
endequation
For the function
beginequation
F(z)=-h^z-1Gamma(1-z)mathcalMleft[H(u) right](z)
endequation
the residues for $z=0,1,2,3$ are
beginalign
left. operatornameresF(z)right|_z=0&=frac1h\
left. operatornameresF(z)right|_z=1&=1\
left. operatornameresF(z)right|_z=2&=frach2left[ lnfrach2+gamma+frac12right]\
left. operatornameresF(z)right|_z=3&=frach^212left[6 lnfrach2+6gamma-1right]
endalign
With $e^-h= 1-h+h^2/2+O(h^3)$,
beginalign
I&sim frac2hM(1-h+frach^22)left[ frac1h+frach2left( lnfrach2+gamma+frac12right)+frach^212left(6 lnfrach2+6gamma-1right)right]\
&sim frac2M+frach^2Mleft( lnfrach2+gamma-frac12 right)
endalign
answered Jul 19 at 15:39
Paul Enta
3,3271925
As noticed, a simple Laplace method cannot be used here, as 2 scales are involved. A uniform asymptotic expansion should be found. Alternatively, in this case, we can recognize a modified Bessel function. Indeed, changing $x=frachMsinh t$, the integral can be written as
beginalign
I&=int_-infty^infty exp(-sqrth^2+M^2x^2) ,dx\
&=frachMint_-infty^infty exp(-hcosh t)cosh t ,dt
endalign
which is proportional to an integral representation of a modified Bessel function (DLMF):
beginequation
K_nuleft(zright)=int_0^inftye^-zcosh tcoshleft(nu tright)%
mathrmdt
endequation
with $nu=1,z=h$,
beginequation
I=frac2hMK_1(h)
endequation
Using the series expansion of the Bessel function near $h=0$ (DLMF),
beginalign
I&sim frac2hMleft[ frac1h+frach2left(lnfrach2 +gamma-frac12right)+ldotsright]\
&simfrac2M+frach^2Mleft(lnfrach2 +gamma-frac12right)+ldots
endalign
EDIT:Another method using the Mellin transform technique
Changing $x=th/M$ and then $u=sqrt1+t^2-1$, the problem is equivalent to find the small $h$ behavior of
beginalign
I&=2frachMint_0^inftyexp(-hsqrt1+t^2),dt\
&=2frachMe^-hint_0^inftyfracu+1sqrtu(u+2)exp(-hu),du
endalign
We have thus to find a Laplace transforms with a small parameter. A classical method which uses the Mellin transform technique is given in (DLMF). Intermediate results are given below, with the help of a CAS. Defining
beginequation
H(u)=fracu+1sqrtu(u+2)
endequation he following behaviors hold:
beginarraylll
H(u)&sim 1+frac12u^2+Oleft( u^-3 right)& text for utoinfty\
&sim O(u^-1/2)& text for uto 0
endarray
and the Mellin transform is
beginequation
mathcalMleft[H(u) right](z)=pi^-1/22^z-1(z-1)Gamma(-z)Gammaleft( z-frac12 right)
endequation
For the function
beginequation
F(z)=-h^z-1Gamma(1-z)mathcalMleft[H(u) right](z)
endequation
the residues for $z=0,1,2,3$ are
beginalign
left. operatornameresF(z)right|_z=0&=frac1h\
left. operatornameresF(z)right|_z=1&=1\
left. operatornameresF(z)right|_z=2&=frach2left[ lnfrach2+gamma+frac12right]\
left. operatornameresF(z)right|_z=3&=frach^212left[6 lnfrach2+6gamma-1right]
endalign
With $e^-h= 1-h+h^2/2+O(h^3)$,
beginalign
I&sim frac2hM(1-h+frach^22)left[ frac1h+frach2left( lnfrach2+gamma+frac12right)+frach^212left(6 lnfrach2+6gamma-1right)right]\
&sim frac2M+frach^2Mleft( lnfrach2+gamma-frac12 right)
endalign
answered Jul 19 at 15:39
Paul Enta
3,3271925
edited Jul 19 at 22:17
answered Jul 19 at 15:39
Paul Enta
3,3271925
answered Jul 19 at 15:39
Paul Enta
3,3271925
answered Jul 19 at 15:39
Paul Enta
3,3271925
3,3271925
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
add a comment |Â
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
This is interesting for this particular problem, but this is really a "toy problem" illustrating the scaling difficulties present in my actual problem of interest (which actually already involves modified Bessel functions). Could you elaborate and/or give a reference about how to obtain a uniform expansion in this situation?
– Ian
Jul 19 at 15:54
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
I included some details to follow the Mellin transform method in the answer.
– Paul Enta
Jul 19 at 22:04
add a comment |Â
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You need $o(h)$ in what sense?
– zhw.
Jul 19 at 16:05
@zhw. I essentially need an estimate for $left. fracdIdh right |_h=0$ where $I$ is the integral considered. Leading order in $M$ is fine, but some reasonable degree of uniformity in $M$ is desirable.
– Ian
Jul 19 at 16:06