Mixing
how to expand $int_{0}^x cos(1/xi)dxi$ like this?
$begingroup$
$$
int_0^x cos(1/xi)dxi
= x+frac{pi}{2}-frac{1}{2x} + frac{2!}{4!3!}frac{1}{x^3}- frac{4!}{6!5!}frac{1}{x^5}cdots
$$
I expanded the Taylor series for $cos(1/xi)$ and integrated that.
but I could not make $pi/2$.
how to make it?
calculus
edited Jan 25 at 5:14
gt6989b
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
$endgroup$
add a comment |
$begingroup$
$$
int_0^x cos(1/xi)dxi
= x+frac{pi}{2}-frac{1}{2x} + frac{2!}{4!3!}frac{1}{x^3}- frac{4!}{6!5!}frac{1}{x^5}cdots
$$
I expanded the Taylor series for $cos(1/xi)$ and integrated that.
but I could not make $pi/2$.
how to make it?
calculus
edited Jan 25 at 5:14
gt6989b
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
$endgroup$
2
$begingroup$
I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
$endgroup$
– Stefan Lafon
Jan 25 at 5:13
$begingroup$
then... Is there any other expansion to make it convergence where x goes to 0?
$endgroup$
– eaststar
Jan 25 at 5:23
1
$begingroup$
I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
$endgroup$
– Robert Israel
Jan 25 at 5:40
$begingroup$
Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
$endgroup$
– Robert Israel
Jan 25 at 5:44
add a comment |
$begingroup$
$$
int_0^x cos(1/xi)dxi
= x+frac{pi}{2}-frac{1}{2x} + frac{2!}{4!3!}frac{1}{x^3}- frac{4!}{6!5!}frac{1}{x^5}cdots
$$
I expanded the Taylor series for $cos(1/xi)$ and integrated that.
but I could not make $pi/2$.
how to make it?
calculus
edited Jan 25 at 5:14
gt6989b
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
$endgroup$
$$
int_0^x cos(1/xi)dxi
= x+frac{pi}{2}-frac{1}{2x} + frac{2!}{4!3!}frac{1}{x^3}- frac{4!}{6!5!}frac{1}{x^5}cdots
$$
I expanded the Taylor series for $cos(1/xi)$ and integrated that.
but I could not make $pi/2$.
how to make it?
calculus
calculus
edited Jan 25 at 5:14
gt6989b
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
edited Jan 25 at 5:14
gt6989b
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
edited Jan 25 at 5:14
gt6989b
35k22557
edited Jan 25 at 5:14
gt6989b
35k22557
edited Jan 25 at 5:14
gt6989b
35k22557
35k22557
asked Jan 25 at 5:08
eaststareaststar
182
asked Jan 25 at 5:08
eaststareaststar
182
asked Jan 25 at 5:08
eaststareaststar
182
182
2
$begingroup$
I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
$endgroup$
– Stefan Lafon
Jan 25 at 5:13
$begingroup$
then... Is there any other expansion to make it convergence where x goes to 0?
$endgroup$
– eaststar
Jan 25 at 5:23
1
$begingroup$
I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
$endgroup$
– Robert Israel
Jan 25 at 5:40
$begingroup$
Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
$endgroup$
– Robert Israel
Jan 25 at 5:44
add a comment |
2
$begingroup$
I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
$endgroup$
– Stefan Lafon
Jan 25 at 5:13
$begingroup$
then... Is there any other expansion to make it convergence where x goes to 0?
$endgroup$
– eaststar
Jan 25 at 5:23
1
$begingroup$
I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
$endgroup$
– Robert Israel
Jan 25 at 5:40
$begingroup$
Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
$endgroup$
– Robert Israel
Jan 25 at 5:44
2
2
$begingroup$
I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
$endgroup$
– Stefan Lafon
Jan 25 at 5:13
$begingroup$
I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
$endgroup$
– Stefan Lafon
Jan 25 at 5:13
$begingroup$
then... Is there any other expansion to make it convergence where x goes to 0?
$endgroup$
– eaststar
Jan 25 at 5:23
$begingroup$
then... Is there any other expansion to make it convergence where x goes to 0?
$endgroup$
– eaststar
Jan 25 at 5:23
1
1
$begingroup$
I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
$endgroup$
– Robert Israel
Jan 25 at 5:40
$begingroup$
I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
$endgroup$
– Robert Israel
Jan 25 at 5:40
$begingroup$
Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
$endgroup$
– Robert Israel
Jan 25 at 5:44
$begingroup$
Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
$endgroup$
– Robert Israel
Jan 25 at 5:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$int_0^xcos(1/xi),dxi=int_{1/x}^inftyfrac{cos t}{t^2},dt$$
at least for $x>0$. Then
$$int_{1/x}^inftyfrac{cos t}{t^2},dt=x-int_{1/x}^infty(1-cos t)frac{dt}{t^2}$$
which looks less than $x$. But
$$int_{1/x}^infty(1-cos t)frac{dt}{t^2}
=int_0^infty(1-cos t)frac{dt}{t^2}
-int_0^{1/x}(1-cos t)frac{dt}{t^2}.$$
The first term here is constant, and is related by integration by parts
to the "sine integral" $int_0^inftyfrac{sin t}t,dt$
and the second integral can be expanded as a Taylor series.
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
add a comment |
$begingroup$
Lord Shark has shown how to derive the expression in question. It is an expansion for $x gg 1$.
Regarding the followup question whether there is an expansion for $0<xll1$: the function is not analytic for small $x$. Nevertheless, we can find an asymptotic expansion for
$$I(x) = int_0^x cos(1/xi)dxi = xint_0^1 cos(1/x t),dt = x operatorname{Re}underbrace{int_0^1 e^{i(x t)^{-1}},dt}_{=J(x)},.$$
In the remaining integral, we perform the substitution $y=1/t$ and obtain
$$J(x) = int_1^infty frac{e^{i y/x}}{y^2} ,dy$$
that is amenable to the saddle point approach. We deform the contour to the integration along the line $1+ieta$, $eta>0$ closed by a quarter circle in the first quadrant of the complex plane. The integral along the circle vanishes as the radius increases and we are left with
$$J(x) = i e^{i/x} int_0^infty frac{e^{-eta/x}}{(1+ i eta)^2} ,dy $$
which is still exact.
We can now expand the denominator of the last expression and obtain
$$ J(x) = i e^{i/x} int_0^infty e^{-eta/x} left( sum_{n=1}^infty n (-i eta)^{n - 1} right) ,dy
sim - e^{i/x} sum_{n=1}^infty n! ,(-i x)^{n}$$
From that we obtain the asymptotic expansion (valid for small $x$)
$$I(x) sim cos(1/x) sum_{m=1}^infty (-1)^{1 + m} (2 m)! ,x^{2 m} + sin(1/x) sum_{m=1}^infty (-1)^m (2m-1)!, x^{2m-1} $$
answered Jan 25 at 6:27
FabianFabian
20k3774
$endgroup$
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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$begingroup$
$$int_0^xcos(1/xi),dxi=int_{1/x}^inftyfrac{cos t}{t^2},dt$$
at least for $x>0$. Then
$$int_{1/x}^inftyfrac{cos t}{t^2},dt=x-int_{1/x}^infty(1-cos t)frac{dt}{t^2}$$
which looks less than $x$. But
$$int_{1/x}^infty(1-cos t)frac{dt}{t^2}
=int_0^infty(1-cos t)frac{dt}{t^2}
-int_0^{1/x}(1-cos t)frac{dt}{t^2}.$$
The first term here is constant, and is related by integration by parts
to the "sine integral" $int_0^inftyfrac{sin t}t,dt$
and the second integral can be expanded as a Taylor series.
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
add a comment |
$begingroup$
$$int_0^xcos(1/xi),dxi=int_{1/x}^inftyfrac{cos t}{t^2},dt$$
at least for $x>0$. Then
$$int_{1/x}^inftyfrac{cos t}{t^2},dt=x-int_{1/x}^infty(1-cos t)frac{dt}{t^2}$$
which looks less than $x$. But
$$int_{1/x}^infty(1-cos t)frac{dt}{t^2}
=int_0^infty(1-cos t)frac{dt}{t^2}
-int_0^{1/x}(1-cos t)frac{dt}{t^2}.$$
The first term here is constant, and is related by integration by parts
to the "sine integral" $int_0^inftyfrac{sin t}t,dt$
and the second integral can be expanded as a Taylor series.
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
add a comment |
$begingroup$
$$int_0^xcos(1/xi),dxi=int_{1/x}^inftyfrac{cos t}{t^2},dt$$
at least for $x>0$. Then
$$int_{1/x}^inftyfrac{cos t}{t^2},dt=x-int_{1/x}^infty(1-cos t)frac{dt}{t^2}$$
which looks less than $x$. But
$$int_{1/x}^infty(1-cos t)frac{dt}{t^2}
=int_0^infty(1-cos t)frac{dt}{t^2}
-int_0^{1/x}(1-cos t)frac{dt}{t^2}.$$
The first term here is constant, and is related by integration by parts
to the "sine integral" $int_0^inftyfrac{sin t}t,dt$
and the second integral can be expanded as a Taylor series.
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$$int_0^xcos(1/xi),dxi=int_{1/x}^inftyfrac{cos t}{t^2},dt$$
at least for $x>0$. Then
$$int_{1/x}^inftyfrac{cos t}{t^2},dt=x-int_{1/x}^infty(1-cos t)frac{dt}{t^2}$$
which looks less than $x$. But
$$int_{1/x}^infty(1-cos t)frac{dt}{t^2}
=int_0^infty(1-cos t)frac{dt}{t^2}
-int_0^{1/x}(1-cos t)frac{dt}{t^2}.$$
The first term here is constant, and is related by integration by parts
to the "sine integral" $int_0^inftyfrac{sin t}t,dt$
and the second integral can be expanded as a Taylor series.
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 5:32
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
add a comment |
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
$begingroup$
Thank you very much!!
$endgroup$
– eaststar
Jan 28 at 8:30
add a comment |
$begingroup$
Lord Shark has shown how to derive the expression in question. It is an expansion for $x gg 1$.
Regarding the followup question whether there is an expansion for $0<xll1$: the function is not analytic for small $x$. Nevertheless, we can find an asymptotic expansion for
$$I(x) = int_0^x cos(1/xi)dxi = xint_0^1 cos(1/x t),dt = x operatorname{Re}underbrace{int_0^1 e^{i(x t)^{-1}},dt}_{=J(x)},.$$
In the remaining integral, we perform the substitution $y=1/t$ and obtain
$$J(x) = int_1^infty frac{e^{i y/x}}{y^2} ,dy$$
that is amenable to the saddle point approach. We deform the contour to the integration along the line $1+ieta$, $eta>0$ closed by a quarter circle in the first quadrant of the complex plane. The integral along the circle vanishes as the radius increases and we are left with
$$J(x) = i e^{i/x} int_0^infty frac{e^{-eta/x}}{(1+ i eta)^2} ,dy $$
which is still exact.
We can now expand the denominator of the last expression and obtain
$$ J(x) = i e^{i/x} int_0^infty e^{-eta/x} left( sum_{n=1}^infty n (-i eta)^{n - 1} right) ,dy
sim - e^{i/x} sum_{n=1}^infty n! ,(-i x)^{n}$$
From that we obtain the asymptotic expansion (valid for small $x$)
$$I(x) sim cos(1/x) sum_{m=1}^infty (-1)^{1 + m} (2 m)! ,x^{2 m} + sin(1/x) sum_{m=1}^infty (-1)^m (2m-1)!, x^{2m-1} $$
answered Jan 25 at 6:27
FabianFabian
20k3774
$endgroup$
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
add a comment |
$begingroup$
Lord Shark has shown how to derive the expression in question. It is an expansion for $x gg 1$.
Regarding the followup question whether there is an expansion for $0<xll1$: the function is not analytic for small $x$. Nevertheless, we can find an asymptotic expansion for
$$I(x) = int_0^x cos(1/xi)dxi = xint_0^1 cos(1/x t),dt = x operatorname{Re}underbrace{int_0^1 e^{i(x t)^{-1}},dt}_{=J(x)},.$$
In the remaining integral, we perform the substitution $y=1/t$ and obtain
$$J(x) = int_1^infty frac{e^{i y/x}}{y^2} ,dy$$
that is amenable to the saddle point approach. We deform the contour to the integration along the line $1+ieta$, $eta>0$ closed by a quarter circle in the first quadrant of the complex plane. The integral along the circle vanishes as the radius increases and we are left with
$$J(x) = i e^{i/x} int_0^infty frac{e^{-eta/x}}{(1+ i eta)^2} ,dy $$
which is still exact.
We can now expand the denominator of the last expression and obtain
$$ J(x) = i e^{i/x} int_0^infty e^{-eta/x} left( sum_{n=1}^infty n (-i eta)^{n - 1} right) ,dy
sim - e^{i/x} sum_{n=1}^infty n! ,(-i x)^{n}$$
From that we obtain the asymptotic expansion (valid for small $x$)
$$I(x) sim cos(1/x) sum_{m=1}^infty (-1)^{1 + m} (2 m)! ,x^{2 m} + sin(1/x) sum_{m=1}^infty (-1)^m (2m-1)!, x^{2m-1} $$
answered Jan 25 at 6:27
FabianFabian
20k3774
$endgroup$
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
add a comment |
$begingroup$
Lord Shark has shown how to derive the expression in question. It is an expansion for $x gg 1$.
Regarding the followup question whether there is an expansion for $0<xll1$: the function is not analytic for small $x$. Nevertheless, we can find an asymptotic expansion for
$$I(x) = int_0^x cos(1/xi)dxi = xint_0^1 cos(1/x t),dt = x operatorname{Re}underbrace{int_0^1 e^{i(x t)^{-1}},dt}_{=J(x)},.$$
In the remaining integral, we perform the substitution $y=1/t$ and obtain
$$J(x) = int_1^infty frac{e^{i y/x}}{y^2} ,dy$$
that is amenable to the saddle point approach. We deform the contour to the integration along the line $1+ieta$, $eta>0$ closed by a quarter circle in the first quadrant of the complex plane. The integral along the circle vanishes as the radius increases and we are left with
$$J(x) = i e^{i/x} int_0^infty frac{e^{-eta/x}}{(1+ i eta)^2} ,dy $$
which is still exact.
We can now expand the denominator of the last expression and obtain
$$ J(x) = i e^{i/x} int_0^infty e^{-eta/x} left( sum_{n=1}^infty n (-i eta)^{n - 1} right) ,dy
sim - e^{i/x} sum_{n=1}^infty n! ,(-i x)^{n}$$
From that we obtain the asymptotic expansion (valid for small $x$)
$$I(x) sim cos(1/x) sum_{m=1}^infty (-1)^{1 + m} (2 m)! ,x^{2 m} + sin(1/x) sum_{m=1}^infty (-1)^m (2m-1)!, x^{2m-1} $$
answered Jan 25 at 6:27
FabianFabian
20k3774
$endgroup$
Lord Shark has shown how to derive the expression in question. It is an expansion for $x gg 1$.
Regarding the followup question whether there is an expansion for $0<xll1$: the function is not analytic for small $x$. Nevertheless, we can find an asymptotic expansion for
$$I(x) = int_0^x cos(1/xi)dxi = xint_0^1 cos(1/x t),dt = x operatorname{Re}underbrace{int_0^1 e^{i(x t)^{-1}},dt}_{=J(x)},.$$
In the remaining integral, we perform the substitution $y=1/t$ and obtain
$$J(x) = int_1^infty frac{e^{i y/x}}{y^2} ,dy$$
that is amenable to the saddle point approach. We deform the contour to the integration along the line $1+ieta$, $eta>0$ closed by a quarter circle in the first quadrant of the complex plane. The integral along the circle vanishes as the radius increases and we are left with
$$J(x) = i e^{i/x} int_0^infty frac{e^{-eta/x}}{(1+ i eta)^2} ,dy $$
which is still exact.
We can now expand the denominator of the last expression and obtain
$$ J(x) = i e^{i/x} int_0^infty e^{-eta/x} left( sum_{n=1}^infty n (-i eta)^{n - 1} right) ,dy
sim - e^{i/x} sum_{n=1}^infty n! ,(-i x)^{n}$$
From that we obtain the asymptotic expansion (valid for small $x$)
$$I(x) sim cos(1/x) sum_{m=1}^infty (-1)^{1 + m} (2 m)! ,x^{2 m} + sin(1/x) sum_{m=1}^infty (-1)^m (2m-1)!, x^{2m-1} $$
answered Jan 25 at 6:27
FabianFabian
20k3774
answered Jan 25 at 6:27
FabianFabian
20k3774
answered Jan 25 at 6:27
FabianFabian
20k3774
answered Jan 25 at 6:27
FabianFabian
20k3774
20k3774
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
add a comment |
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
$begingroup$
You give me the desire to study!! Thank you~~
$endgroup$
– eaststar
Jan 28 at 8:32
add a comment |
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Required, but never shown
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I initially thought this was some expansion around $x=0$. But the left-hand side goes to zeros as $xrightarrow 0$ while the right-hand side has divergent terms. Where is that expansion valid?
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– Stefan Lafon
Jan 25 at 5:13
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then... Is there any other expansion to make it convergence where x goes to 0?
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– eaststar
Jan 25 at 5:23
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I think you have some signs wrong. According to Maple it should be (for $x > 0$) $$ x - frac{pi}{2} + frac{1}{2x} - frac{1}{72 x^3} + frac{1}{3600 x^5} ldots $$
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– Robert Israel
Jan 25 at 5:40
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Note that since $cos(1/xi) le 1$ the integral must be less than $x$. Your formula is greater than $x$ for large $x$.
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– Robert Israel
Jan 25 at 5:44