Asymptotic expansion of modified Laplace type integral
Clash Royale CLAN TAG#URR8PPP
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I am looking for the asymptotic expansion for $tau rightarrow0$ of the following integral :
$I(tau):=int_0^1sqrt1-wexpleft[-fracleft(ln S-barxleft(tau wright)right)^22tauleft(1-wright)sigma^2right]rm d w$
where $bar x(0)=ln K<ln S$ and $bar x(tau) = ln K - sigmasqrt-tau ln tau+o(sqrt-tau ln tau)$ and $bar x$ is a non-increasing function defined on $mathbb R+$ and $barx'(tau)rightarrowinfty$ as $tau rightarrow 0$.
The problem is that $I$ is not directly a Laplace type integral as there is still $tau$ in the function $bar x$. I am pretty confident that the leading behaviour of the $I$ is determined by the values of the integrand near the point w=0 as the term in the exponential will be maximized at that point. Can we therefore replace the exponential by its equivalent : $expleft[-fracleft(ln S-ln Kright)^22tauleft(1-wright)sigma^2right]$ ?
If anyone has any idea how to solve such problem please help.
Thank you !
real-analysis integration limits asymptotics laplace-method
asked Sep 6 at 10:41
flo3299
164
add a comment |Â
up vote
0
down vote
favorite
I am looking for the asymptotic expansion for $tau rightarrow0$ of the following integral :
$I(tau):=int_0^1sqrt1-wexpleft[-fracleft(ln S-barxleft(tau wright)right)^22tauleft(1-wright)sigma^2right]rm d w$
where $bar x(0)=ln K<ln S$ and $bar x(tau) = ln K - sigmasqrt-tau ln tau+o(sqrt-tau ln tau)$ and $bar x$ is a non-increasing function defined on $mathbb R+$ and $barx'(tau)rightarrowinfty$ as $tau rightarrow 0$.
The problem is that $I$ is not directly a Laplace type integral as there is still $tau$ in the function $bar x$. I am pretty confident that the leading behaviour of the $I$ is determined by the values of the integrand near the point w=0 as the term in the exponential will be maximized at that point. Can we therefore replace the exponential by its equivalent : $expleft[-fracleft(ln S-ln Kright)^22tauleft(1-wright)sigma^2right]$ ?
If anyone has any idea how to solve such problem please help.
Thank you !
real-analysis integration limits asymptotics laplace-method
asked Sep 6 at 10:41
flo3299
164
anyone, maybe ?
– flo3299
Sep 11 at 16:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am looking for the asymptotic expansion for $tau rightarrow0$ of the following integral :
$I(tau):=int_0^1sqrt1-wexpleft[-fracleft(ln S-barxleft(tau wright)right)^22tauleft(1-wright)sigma^2right]rm d w$
where $bar x(0)=ln K<ln S$ and $bar x(tau) = ln K - sigmasqrt-tau ln tau+o(sqrt-tau ln tau)$ and $bar x$ is a non-increasing function defined on $mathbb R+$ and $barx'(tau)rightarrowinfty$ as $tau rightarrow 0$.
The problem is that $I$ is not directly a Laplace type integral as there is still $tau$ in the function $bar x$. I am pretty confident that the leading behaviour of the $I$ is determined by the values of the integrand near the point w=0 as the term in the exponential will be maximized at that point. Can we therefore replace the exponential by its equivalent : $expleft[-fracleft(ln S-ln Kright)^22tauleft(1-wright)sigma^2right]$ ?
If anyone has any idea how to solve such problem please help.
Thank you !
real-analysis integration limits asymptotics laplace-method
asked Sep 6 at 10:41
flo3299
164
I am looking for the asymptotic expansion for $tau rightarrow0$ of the following integral :
$I(tau):=int_0^1sqrt1-wexpleft[-fracleft(ln S-barxleft(tau wright)right)^22tauleft(1-wright)sigma^2right]rm d w$
where $bar x(0)=ln K<ln S$ and $bar x(tau) = ln K - sigmasqrt-tau ln tau+o(sqrt-tau ln tau)$ and $bar x$ is a non-increasing function defined on $mathbb R+$ and $barx'(tau)rightarrowinfty$ as $tau rightarrow 0$.
The problem is that $I$ is not directly a Laplace type integral as there is still $tau$ in the function $bar x$. I am pretty confident that the leading behaviour of the $I$ is determined by the values of the integrand near the point w=0 as the term in the exponential will be maximized at that point. Can we therefore replace the exponential by its equivalent : $expleft[-fracleft(ln S-ln Kright)^22tauleft(1-wright)sigma^2right]$ ?
If anyone has any idea how to solve such problem please help.
Thank you !
real-analysis integration limits asymptotics laplace-method
real-analysis integration limits asymptotics laplace-method
asked Sep 6 at 10:41
flo3299
164
asked Sep 6 at 10:41
flo3299
164
asked Sep 6 at 10:41
flo3299
164
asked Sep 6 at 10:41
flo3299
164
asked Sep 6 at 10:41
flo3299
164
164
anyone, maybe ?
– flo3299
Sep 11 at 16:30
add a comment |Â
anyone, maybe ?
– flo3299
Sep 11 at 16:30
anyone, maybe ?
– flo3299
Sep 11 at 16:30
anyone, maybe ?
– flo3299
Sep 11 at 16:30
add a comment |Â
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anyone, maybe ?
– flo3299
Sep 11 at 16:30