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Compute $lim_xto 0^+int_2x^3xfracsin(t)sinh^2(t)dt$.
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How can I compute $$lim_xto 0^+int_2x^3xfracsin(t)sinh^2(t)dt ?$$
Suppose $x<pi$. I tried using DCT since $$left|fracsin(t)sinh^2(t)boldsymbol 1_[2x,3x(t)right|leq fracsin(t)sin^2(t),$$
but the function on the RHS is not integrable... It should have a trick.
real-analysis integration limits
asked Jul 18 at 9:42
MathBeginner
707312
add a comment |Â
up vote
4
down vote
favorite
How can I compute $$lim_xto 0^+int_2x^3xfracsin(t)sinh^2(t)dt ?$$
Suppose $x<pi$. I tried using DCT since $$left|fracsin(t)sinh^2(t)boldsymbol 1_[2x,3x(t)right|leq fracsin(t)sin^2(t),$$
but the function on the RHS is not integrable... It should have a trick.
real-analysis integration limits
asked Jul 18 at 9:42
MathBeginner
707312
The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
How can I compute $$lim_xto 0^+int_2x^3xfracsin(t)sinh^2(t)dt ?$$
Suppose $x<pi$. I tried using DCT since $$left|fracsin(t)sinh^2(t)boldsymbol 1_[2x,3x(t)right|leq fracsin(t)sin^2(t),$$
but the function on the RHS is not integrable... It should have a trick.
real-analysis integration limits
asked Jul 18 at 9:42
MathBeginner
707312
How can I compute $$lim_xto 0^+int_2x^3xfracsin(t)sinh^2(t)dt ?$$
Suppose $x<pi$. I tried using DCT since $$left|fracsin(t)sinh^2(t)boldsymbol 1_[2x,3x(t)right|leq fracsin(t)sin^2(t),$$
but the function on the RHS is not integrable... It should have a trick.
real-analysis integration limits
asked Jul 18 at 9:42
MathBeginner
707312
asked Jul 18 at 9:42
MathBeginner
707312
asked Jul 18 at 9:42
MathBeginner
707312
asked Jul 18 at 9:42
MathBeginner
707312
707312
The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36
add a comment |Â
The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36
The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36
add a comment |Â
4 Answers
4
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You have$$lim_tto0^+fracsin(t)sinh^2(t)-frac1t=lim_tto0^+fractsin(t)-sinh^2(t)tsinh^2(t)=0$$and thereforebeginalignlim_xto0^+int_2x^3xfracsin(t)sinh^2(t),mathrm dt&=lim_xto0^+left(int_2x^3xfracsin(t)sinh^2(t)-frac1t,mathrm dtright)+lim_xto0^+int_2x^3xfracmathrm dtt\&=lim_xto0^+bigl(log(3x)-log(2x)bigr)\&=logleft(frac32right).endalign
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
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up vote
1
down vote
The Taylor development of the smooth even function
$$fractsin tsinh^2t$$
is of the form
$$1-fract^22+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2lefractsin tsinh^2tle1.$$
(In fact $a=dfrac12$ works.)
This implies
$$frac1t-atlefracsin tsinh^2tlefrac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$logfrac32-axle Ile logfrac32.$$
answered Jul 18 at 10:59
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
Put $ f(t)=dfractsin(t)sinh^2(t) $. Then
$lim_tto 0^+f(t) = 1$.
From the Second mean value theorem for definite integrals we get
beginequation*
int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = int_2x^3xf(t)dfrac1t, mathrmdt = f(xi)int_2x^3xdfrac1t, mathrmdt = f(xi)logleft(dfrac32right)
endequation*
where $ 2x<xi <3x$. Consequently
beginequation*
lim_tto 0^+int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = logleft(dfrac32right).
endequation*
answered Jul 18 at 11:38
JanG
2,572412
add a comment |Â
up vote
0
down vote
You could also use series expansion of the integrand. This would give
$$fracsin(t)sinh^2(t)=fract-fract^36+fract^5120+Oleft(t^7right) left(t+fract^36+fract^5120+Oleft(t^7right) right)^2 =fract-fract^36+fract^5120+Oleft(t^7right) t^2+fract^43+frac2 t^645+Oleft(t^8right) =frac1t-fract2+frac47 t^3360+Oleft(t^5right)$$ Now, integrate, use the bounds and so on.
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
You have$$lim_tto0^+fracsin(t)sinh^2(t)-frac1t=lim_tto0^+fractsin(t)-sinh^2(t)tsinh^2(t)=0$$and thereforebeginalignlim_xto0^+int_2x^3xfracsin(t)sinh^2(t),mathrm dt&=lim_xto0^+left(int_2x^3xfracsin(t)sinh^2(t)-frac1t,mathrm dtright)+lim_xto0^+int_2x^3xfracmathrm dtt\&=lim_xto0^+bigl(log(3x)-log(2x)bigr)\&=logleft(frac32right).endalign
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
add a comment |Â
up vote
5
down vote
You have$$lim_tto0^+fracsin(t)sinh^2(t)-frac1t=lim_tto0^+fractsin(t)-sinh^2(t)tsinh^2(t)=0$$and thereforebeginalignlim_xto0^+int_2x^3xfracsin(t)sinh^2(t),mathrm dt&=lim_xto0^+left(int_2x^3xfracsin(t)sinh^2(t)-frac1t,mathrm dtright)+lim_xto0^+int_2x^3xfracmathrm dtt\&=lim_xto0^+bigl(log(3x)-log(2x)bigr)\&=logleft(frac32right).endalign
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You have$$lim_tto0^+fracsin(t)sinh^2(t)-frac1t=lim_tto0^+fractsin(t)-sinh^2(t)tsinh^2(t)=0$$and thereforebeginalignlim_xto0^+int_2x^3xfracsin(t)sinh^2(t),mathrm dt&=lim_xto0^+left(int_2x^3xfracsin(t)sinh^2(t)-frac1t,mathrm dtright)+lim_xto0^+int_2x^3xfracmathrm dtt\&=lim_xto0^+bigl(log(3x)-log(2x)bigr)\&=logleft(frac32right).endalign
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
You have$$lim_tto0^+fracsin(t)sinh^2(t)-frac1t=lim_tto0^+fractsin(t)-sinh^2(t)tsinh^2(t)=0$$and thereforebeginalignlim_xto0^+int_2x^3xfracsin(t)sinh^2(t),mathrm dt&=lim_xto0^+left(int_2x^3xfracsin(t)sinh^2(t)-frac1t,mathrm dtright)+lim_xto0^+int_2x^3xfracmathrm dtt\&=lim_xto0^+bigl(log(3x)-log(2x)bigr)\&=logleft(frac32right).endalign
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
answered Jul 18 at 9:50
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
1
down vote
The Taylor development of the smooth even function
$$fractsin tsinh^2t$$
is of the form
$$1-fract^22+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2lefractsin tsinh^2tle1.$$
(In fact $a=dfrac12$ works.)
This implies
$$frac1t-atlefracsin tsinh^2tlefrac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$logfrac32-axle Ile logfrac32.$$
answered Jul 18 at 10:59
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
The Taylor development of the smooth even function
$$fractsin tsinh^2t$$
is of the form
$$1-fract^22+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2lefractsin tsinh^2tle1.$$
(In fact $a=dfrac12$ works.)
This implies
$$frac1t-atlefracsin tsinh^2tlefrac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$logfrac32-axle Ile logfrac32.$$
answered Jul 18 at 10:59
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The Taylor development of the smooth even function
$$fractsin tsinh^2t$$
is of the form
$$1-fract^22+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2lefractsin tsinh^2tle1.$$
(In fact $a=dfrac12$ works.)
This implies
$$frac1t-atlefracsin tsinh^2tlefrac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$logfrac32-axle Ile logfrac32.$$
answered Jul 18 at 10:59
Yves Daoust
111k665204
The Taylor development of the smooth even function
$$fractsin tsinh^2t$$
is of the form
$$1-fract^22+o(t^2)$$
so that for small $t$ there is certainly a finite $a$ such that
$$1-at^2lefractsin tsinh^2tle1.$$
(In fact $a=dfrac12$ works.)
This implies
$$frac1t-atlefracsin tsinh^2tlefrac1t,$$
and integrating between $2x$ and $3x$ ($<t$),
$$logfrac32-axle Ile logfrac32.$$
answered Jul 18 at 10:59
Yves Daoust
111k665204
edited Jul 18 at 11:13
answered Jul 18 at 10:59
Yves Daoust
111k665204
answered Jul 18 at 10:59
Yves Daoust
111k665204
answered Jul 18 at 10:59
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
up vote
1
down vote
Put $ f(t)=dfractsin(t)sinh^2(t) $. Then
$lim_tto 0^+f(t) = 1$.
From the Second mean value theorem for definite integrals we get
beginequation*
int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = int_2x^3xf(t)dfrac1t, mathrmdt = f(xi)int_2x^3xdfrac1t, mathrmdt = f(xi)logleft(dfrac32right)
endequation*
where $ 2x<xi <3x$. Consequently
beginequation*
lim_tto 0^+int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = logleft(dfrac32right).
endequation*
answered Jul 18 at 11:38
JanG
2,572412
add a comment |Â
up vote
1
down vote
Put $ f(t)=dfractsin(t)sinh^2(t) $. Then
$lim_tto 0^+f(t) = 1$.
From the Second mean value theorem for definite integrals we get
beginequation*
int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = int_2x^3xf(t)dfrac1t, mathrmdt = f(xi)int_2x^3xdfrac1t, mathrmdt = f(xi)logleft(dfrac32right)
endequation*
where $ 2x<xi <3x$. Consequently
beginequation*
lim_tto 0^+int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = logleft(dfrac32right).
endequation*
answered Jul 18 at 11:38
JanG
2,572412
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Put $ f(t)=dfractsin(t)sinh^2(t) $. Then
$lim_tto 0^+f(t) = 1$.
From the Second mean value theorem for definite integrals we get
beginequation*
int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = int_2x^3xf(t)dfrac1t, mathrmdt = f(xi)int_2x^3xdfrac1t, mathrmdt = f(xi)logleft(dfrac32right)
endequation*
where $ 2x<xi <3x$. Consequently
beginequation*
lim_tto 0^+int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = logleft(dfrac32right).
endequation*
answered Jul 18 at 11:38
JanG
2,572412
Put $ f(t)=dfractsin(t)sinh^2(t) $. Then
$lim_tto 0^+f(t) = 1$.
From the Second mean value theorem for definite integrals we get
beginequation*
int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = int_2x^3xf(t)dfrac1t, mathrmdt = f(xi)int_2x^3xdfrac1t, mathrmdt = f(xi)logleft(dfrac32right)
endequation*
where $ 2x<xi <3x$. Consequently
beginequation*
lim_tto 0^+int_2x^3xdfracsin(t)sinh^2(t), mathrmdt = logleft(dfrac32right).
endequation*
answered Jul 18 at 11:38
JanG
2,572412
answered Jul 18 at 11:38
JanG
2,572412
answered Jul 18 at 11:38
JanG
2,572412
answered Jul 18 at 11:38
JanG
2,572412
2,572412
add a comment |Â
add a comment |Â
up vote
0
down vote
You could also use series expansion of the integrand. This would give
$$fracsin(t)sinh^2(t)=fract-fract^36+fract^5120+Oleft(t^7right) left(t+fract^36+fract^5120+Oleft(t^7right) right)^2 =fract-fract^36+fract^5120+Oleft(t^7right) t^2+fract^43+frac2 t^645+Oleft(t^8right) =frac1t-fract2+frac47 t^3360+Oleft(t^5right)$$ Now, integrate, use the bounds and so on.
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
add a comment |Â
up vote
0
down vote
You could also use series expansion of the integrand. This would give
$$fracsin(t)sinh^2(t)=fract-fract^36+fract^5120+Oleft(t^7right) left(t+fract^36+fract^5120+Oleft(t^7right) right)^2 =fract-fract^36+fract^5120+Oleft(t^7right) t^2+fract^43+frac2 t^645+Oleft(t^8right) =frac1t-fract2+frac47 t^3360+Oleft(t^5right)$$ Now, integrate, use the bounds and so on.
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You could also use series expansion of the integrand. This would give
$$fracsin(t)sinh^2(t)=fract-fract^36+fract^5120+Oleft(t^7right) left(t+fract^36+fract^5120+Oleft(t^7right) right)^2 =fract-fract^36+fract^5120+Oleft(t^7right) t^2+fract^43+frac2 t^645+Oleft(t^8right) =frac1t-fract2+frac47 t^3360+Oleft(t^5right)$$ Now, integrate, use the bounds and so on.
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
You could also use series expansion of the integrand. This would give
$$fracsin(t)sinh^2(t)=fract-fract^36+fract^5120+Oleft(t^7right) left(t+fract^36+fract^5120+Oleft(t^7right) right)^2 =fract-fract^36+fract^5120+Oleft(t^7right) t^2+fract^43+frac2 t^645+Oleft(t^8right) =frac1t-fract2+frac47 t^3360+Oleft(t^5right)$$ Now, integrate, use the bounds and so on.
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
answered Jul 18 at 10:17
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
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The integrand is asymptotic to $1/t$.
– Yves Daoust
Jul 18 at 10:20
For the fun of it, try numerical integration with $x=frac 14$; you should get $approx 0.335004$ while, using the expansion, you should get $approx 0.335627$.
– Claude Leibovici
Jul 18 at 10:36