Use Laplace method to solve $I(x)= int^{frac{pi}{2}}_0 (1+t^2)e^{-xcosh t}dt$
Consider the integral
$$I(x)= int^{frac{pi}{2}}_0 (1+t^2)e^{-xcosh t}dt$$
Use Laplace's method to show that
$$I(x) = sqrt{frac{pi}{2}}e^{-x}+cfrac{e^{-x}}{x}, as space x to infty$$
where $c$ is a constant you must determine.
Where I have got to so far:
let $phi(t)=-xcosh t$
$phi'(t)=xsinh t$
$phi''(t)=xcosh t$
$$I(x) sim int_0^{epsilon}(1+t^2)e^{-xcosh t}dt space + space int_{epsilon}^{frac{pi}{2}}(1+t^2)e^{-xcosh t}dt$$
integration laplace-method
edited Dec 12 '18 at 15:30
Jon
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
add a comment |
Consider the integral
$$I(x)= int^{frac{pi}{2}}_0 (1+t^2)e^{-xcosh t}dt$$
Use Laplace's method to show that
$$I(x) = sqrt{frac{pi}{2}}e^{-x}+cfrac{e^{-x}}{x}, as space x to infty$$
where $c$ is a constant you must determine.
Where I have got to so far:
let $phi(t)=-xcosh t$
$phi'(t)=xsinh t$
$phi''(t)=xcosh t$
$$I(x) sim int_0^{epsilon}(1+t^2)e^{-xcosh t}dt space + space int_{epsilon}^{frac{pi}{2}}(1+t^2)e^{-xcosh t}dt$$
integration laplace-method
edited Dec 12 '18 at 15:30
Jon
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
add a comment |
Consider the integral
$$I(x)= int^{frac{pi}{2}}_0 (1+t^2)e^{-xcosh t}dt$$
Use Laplace's method to show that
$$I(x) = sqrt{frac{pi}{2}}e^{-x}+cfrac{e^{-x}}{x}, as space x to infty$$
where $c$ is a constant you must determine.
Where I have got to so far:
let $phi(t)=-xcosh t$
$phi'(t)=xsinh t$
$phi''(t)=xcosh t$
$$I(x) sim int_0^{epsilon}(1+t^2)e^{-xcosh t}dt space + space int_{epsilon}^{frac{pi}{2}}(1+t^2)e^{-xcosh t}dt$$
integration laplace-method
edited Dec 12 '18 at 15:30
Jon
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
Consider the integral
$$I(x)= int^{frac{pi}{2}}_0 (1+t^2)e^{-xcosh t}dt$$
Use Laplace's method to show that
$$I(x) = sqrt{frac{pi}{2}}e^{-x}+cfrac{e^{-x}}{x}, as space x to infty$$
where $c$ is a constant you must determine.
Where I have got to so far:
let $phi(t)=-xcosh t$
$phi'(t)=xsinh t$
$phi''(t)=xcosh t$
$$I(x) sim int_0^{epsilon}(1+t^2)e^{-xcosh t}dt space + space int_{epsilon}^{frac{pi}{2}}(1+t^2)e^{-xcosh t}dt$$
integration laplace-method
integration laplace-method
edited Dec 12 '18 at 15:30
Jon
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
edited Dec 12 '18 at 15:30
Jon
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
edited Dec 12 '18 at 15:30
Jon
4,39011122
edited Dec 12 '18 at 15:30
Jon
4,39011122
edited Dec 12 '18 at 15:30
Jon
4,39011122
4,39011122
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
asked Dec 12 '18 at 15:14
Ben JonesBen Jones
19111
19111
add a comment |
add a comment |
1 Answer
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A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-xcosh(t) + log(1+t^2)$ about its maximum at $t=0$ I get
$$
I(x) sim int_0^epsilon {rm e}^{-x-frac{x-2}{2}, t^2} , {rm d}t = sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , {rm erf}left( sqrt{frac{x-2}{2}} , epsilon right) sim sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , .
$$
If I'm not mistaken I think the next term should come from
$$
int_0^{frac{pi}{2}} left{ {rm e}^{-x-frac{x-2}{2},t^2} - (1+t^2) , {rm e}^{-xcosh(t)} right} {rm d}t sim frac{sqrt{2pi}}{16 , x^{3/2}}, {rm e}^{-x} , ,
$$
so we would have
$$
int_0^{frac{pi}{2}} (1+t^2) , {rm e}^{-x cosh(t)} , {rm d}t sim sqrt{frac{pi}{2x}} , {rm e}^{-x} + frac{7sqrt{2pi}}{16 , x^{3/2}} , {rm e}^{-x} , .
$$
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
add a comment |
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1 Answer
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1 Answer
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active
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A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-xcosh(t) + log(1+t^2)$ about its maximum at $t=0$ I get
$$
I(x) sim int_0^epsilon {rm e}^{-x-frac{x-2}{2}, t^2} , {rm d}t = sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , {rm erf}left( sqrt{frac{x-2}{2}} , epsilon right) sim sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , .
$$
If I'm not mistaken I think the next term should come from
$$
int_0^{frac{pi}{2}} left{ {rm e}^{-x-frac{x-2}{2},t^2} - (1+t^2) , {rm e}^{-xcosh(t)} right} {rm d}t sim frac{sqrt{2pi}}{16 , x^{3/2}}, {rm e}^{-x} , ,
$$
so we would have
$$
int_0^{frac{pi}{2}} (1+t^2) , {rm e}^{-x cosh(t)} , {rm d}t sim sqrt{frac{pi}{2x}} , {rm e}^{-x} + frac{7sqrt{2pi}}{16 , x^{3/2}} , {rm e}^{-x} , .
$$
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
add a comment |
A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-xcosh(t) + log(1+t^2)$ about its maximum at $t=0$ I get
$$
I(x) sim int_0^epsilon {rm e}^{-x-frac{x-2}{2}, t^2} , {rm d}t = sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , {rm erf}left( sqrt{frac{x-2}{2}} , epsilon right) sim sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , .
$$
If I'm not mistaken I think the next term should come from
$$
int_0^{frac{pi}{2}} left{ {rm e}^{-x-frac{x-2}{2},t^2} - (1+t^2) , {rm e}^{-xcosh(t)} right} {rm d}t sim frac{sqrt{2pi}}{16 , x^{3/2}}, {rm e}^{-x} , ,
$$
so we would have
$$
int_0^{frac{pi}{2}} (1+t^2) , {rm e}^{-x cosh(t)} , {rm d}t sim sqrt{frac{pi}{2x}} , {rm e}^{-x} + frac{7sqrt{2pi}}{16 , x^{3/2}} , {rm e}^{-x} , .
$$
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
add a comment |
A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-xcosh(t) + log(1+t^2)$ about its maximum at $t=0$ I get
$$
I(x) sim int_0^epsilon {rm e}^{-x-frac{x-2}{2}, t^2} , {rm d}t = sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , {rm erf}left( sqrt{frac{x-2}{2}} , epsilon right) sim sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , .
$$
If I'm not mistaken I think the next term should come from
$$
int_0^{frac{pi}{2}} left{ {rm e}^{-x-frac{x-2}{2},t^2} - (1+t^2) , {rm e}^{-xcosh(t)} right} {rm d}t sim frac{sqrt{2pi}}{16 , x^{3/2}}, {rm e}^{-x} , ,
$$
so we would have
$$
int_0^{frac{pi}{2}} (1+t^2) , {rm e}^{-x cosh(t)} , {rm d}t sim sqrt{frac{pi}{2x}} , {rm e}^{-x} + frac{7sqrt{2pi}}{16 , x^{3/2}} , {rm e}^{-x} , .
$$
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
A bit too long of a comment:
Are you sure about the claim? If I expand the exponent $-xcosh(t) + log(1+t^2)$ about its maximum at $t=0$ I get
$$
I(x) sim int_0^epsilon {rm e}^{-x-frac{x-2}{2}, t^2} , {rm d}t = sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , {rm erf}left( sqrt{frac{x-2}{2}} , epsilon right) sim sqrt{frac{pi/2}{x-2}} , {rm e}^{-x} , .
$$
If I'm not mistaken I think the next term should come from
$$
int_0^{frac{pi}{2}} left{ {rm e}^{-x-frac{x-2}{2},t^2} - (1+t^2) , {rm e}^{-xcosh(t)} right} {rm d}t sim frac{sqrt{2pi}}{16 , x^{3/2}}, {rm e}^{-x} , ,
$$
so we would have
$$
int_0^{frac{pi}{2}} (1+t^2) , {rm e}^{-x cosh(t)} , {rm d}t sim sqrt{frac{pi}{2x}} , {rm e}^{-x} + frac{7sqrt{2pi}}{16 , x^{3/2}} , {rm e}^{-x} , .
$$
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
edited Dec 12 '18 at 19:54
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
answered Dec 12 '18 at 16:57
DigerDiger
1,6021413
1,6021413
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
add a comment |
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
For "large" $x$, $x-2sim x$. So, you have recovered the first term in the expansion that corrects the first term provided in the OP.
– Mark Viola
Dec 12 '18 at 17:52
add a comment |
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