Mixing
large x and small x expansion for gamma-like function
up vote
2
down vote
favorite
Find two approximations for the integral ($x>0$)
begin{equation}
I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta
end{equation}
one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).
For the small x expansion, I tried using a Maclauren series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer. I know this is likely related to the Gamma function, but am fairly novice at problems of this form. Any help deeply appreciated!
sequences-and-series complex-analysis asymptotics gamma-function
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 1 more comment
up vote
2
down vote
favorite
Find two approximations for the integral ($x>0$)
begin{equation}
I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta
end{equation}
one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).
For the small x expansion, I tried using a Maclauren series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer. I know this is likely related to the Gamma function, but am fairly novice at problems of this form. Any help deeply appreciated!
sequences-and-series complex-analysis asymptotics gamma-function
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
1
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find two approximations for the integral ($x>0$)
begin{equation}
I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta
end{equation}
one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).
For the small x expansion, I tried using a Maclauren series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer. I know this is likely related to the Gamma function, but am fairly novice at problems of this form. Any help deeply appreciated!
sequences-and-series complex-analysis asymptotics gamma-function
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Find two approximations for the integral ($x>0$)
begin{equation}
I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta
end{equation}
one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).
For the small x expansion, I tried using a Maclauren series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer. I know this is likely related to the Gamma function, but am fairly novice at problems of this form. Any help deeply appreciated!
sequences-and-series complex-analysis asymptotics gamma-function
sequences-and-series complex-analysis asymptotics gamma-function
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
niagarajohn
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
niagarajohn
184
asked 2 days ago
niagarajohn
184
184
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
niagarajohn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
1
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday
|
show 1 more comment
1
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
1
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday
1
1
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
1
1
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
0
down vote
How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$
EDIT:
taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$
answered 2 days ago
Henry Lee
1,588117
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
add a comment |
up vote
0
down vote
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.
Let's transform the integral first:
$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$
Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).
Using the notation from the article, we have:
$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$
Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.
Then we can represent:
$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
answered 20 hours ago
Yuriy S
15.5k333114
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$
EDIT:
taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$
answered 2 days ago
Henry Lee
1,588117
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
add a comment |
up vote
0
down vote
How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$
EDIT:
taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$
answered 2 days ago
Henry Lee
1,588117
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$
EDIT:
taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$
answered 2 days ago
Henry Lee
1,588117
How about this:
$$I(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{xcos^2theta}dtheta=frac{1}{2pi}int_{-pi/2}^{pi/2}e^{x(1-sin^2theta)}dtheta$$
$$theta=pi/2-phi$$
$$dtheta=-dphi$$
$$I(x)=frac{1}{2pi}int_0^pi e^{xsin^2phi}dphi$$
EDIT:
taking the other approach, I have:
$$I'(x)=frac{1}{2pi}int_{-pi/2}^{pi/2}cos^2theta e^{xcos^2theta}dtheta$$
then using $u=costheta$ you can obtain:
$$I'(x)=frac1piint_0^1frac{u^2}{sqrt{1-u^2}}e^{xu^2}du$$
now using $v=sqrt{1-u^2}$ we can get:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=frac1piint_0^1sqrt{1-v^2}e^{-(sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=frac{1}{2pi x^2}int_0^xbeta e^beta dbeta$$
so:
$$I(x)=frac1{2pi}iintfrac{(x-1)e^x+1}{x^2}dxdx$$
answered 2 days ago
Henry Lee
1,588117
edited yesterday
answered 2 days ago
Henry Lee
1,588117
answered 2 days ago
Henry Lee
1,588117
answered 2 days ago
Henry Lee
1,588117
1,588117
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
add a comment |
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
1
1
How does this help find the asymptotic?
– Yuriy S
2 days ago
How does this help find the asymptotic?
– Yuriy S
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
You could try to get a term for $I^2(x)$ then use polar coordinates
– Henry Lee
2 days ago
1
1
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
I think you are onto something here. I've seen large and small x expansions for the erf function and this could be the trick to get it into that form. Will take a look!
– niagarajohn
2 days ago
^That didn't work...
– niagarajohn
yesterday
^That didn't work...
– niagarajohn
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
Ok try and find $I’(x)$ then make a substitution, find a term for $I’(x)$ in terms of x then integrate this
– Henry Lee
yesterday
add a comment |
up vote
0
down vote
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.
Let's transform the integral first:
$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$
Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).
Using the notation from the article, we have:
$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$
Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.
Then we can represent:
$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
answered 20 hours ago
Yuriy S
15.5k333114
add a comment |
up vote
0
down vote
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.
Let's transform the integral first:
$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$
Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).
Using the notation from the article, we have:
$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$
Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.
Then we can represent:
$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
answered 20 hours ago
Yuriy S
15.5k333114
add a comment |
up vote
0
down vote
up vote
0
down vote
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.
Let's transform the integral first:
$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$
Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).
Using the notation from the article, we have:
$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$
Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.
Then we can represent:
$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
answered 20 hours ago
Yuriy S
15.5k333114
For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.
Let's transform the integral first:
$$ I(x) = frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta =frac{1}{pi}int_0^{frac{pi}{2}}
e^{x cos^2(theta)}dtheta=frac{e^{x}}{pi} int_0^{frac{pi}{2}}
e^{-x sin^2(theta)}dtheta= \ =frac{e^{x}}{pi} int_0^1
frac{e^{-x s^2}}{sqrt{1-s^2}} d s=frac{e^{x}}{2pi} int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t$$
Now for the latter integral, the main contribution for $x to infty$ will be given by $t to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).
Using the notation from the article, we have:
$$phi(t)=frac{1}{sqrt{t}sqrt{1-t}}=t^{-1/2} g(t)$$
Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $int_0^1 |phi(t)| dt=pi<infty$. So the lemma conditions are satisfied.
Then we can represent:
$$int_0^1
frac{e^{-x t}}{sqrt{t}sqrt{1-t}} d t asymp sum_{n=0}^infty frac{Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$
answered 20 hours ago
Yuriy S
15.5k333114
edited 19 hours ago
answered 20 hours ago
Yuriy S
15.5k333114
answered 20 hours ago
Yuriy S
15.5k333114
answered 20 hours ago
Yuriy S
15.5k333114
15.5k333114
add a comment |
add a comment |
niagarajohn is a new contributor. Be nice, and check out our Code of Conduct.
niagarajohn is a new contributor. Be nice, and check out our Code of Conduct.
niagarajohn is a new contributor. Be nice, and check out our Code of Conduct.
niagarajohn is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002851%2flarge-x-and-small-x-expansion-for-gamma-like-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
For large $x$ check out en.wikipedia.org/wiki/Laplace%27s_method
– Yuriy S
2 days ago
Thanks @Yuriy. I haven't been exposed to this method, so I think I'd like to solve using asymptotic series. Still working through a few trials, but am interested in easier ways to solve problems like these outside of my current situation.
– niagarajohn
2 days ago
1
Laplace's method is the standard one for such situations, and it's really simple. But of course, you can choose to try some other way
– Yuriy S
2 days ago
Seconding @Yuriy's suggestion.
– Antonio Vargas
2 days ago
My attempt for small x: begin{align} I(x) &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} e^{x cos^2(theta)}dtheta\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + xcos^2{theta} + order{x^2}]\ &= frac{1}{2pi}int_{-frac{pi}{2}}^{frac{pi}{2}} dtheta [1 + frac{x}{2}(1+cos{2theta}) + order{x^2}]\ &approx frac{1}{2pi} [pi + frac{x}{2}(pi)] end{align} Therefore, begin{align} &boxed{ I(x)approx frac{1}{2} + frac{x}{4} text{for small $x$}} end{align}
– niagarajohn
yesterday