Mixing
Asymptotics for $log$ integral
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up vote
1
down vote
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So I have the following integral:
$$
int_k - 1/2^k + 1/2 log t mathrmdt
$$
And I would like to show it can be written as:
$$
int_k - 1/2^k + 1/2 log t mathrmdt = log k + mathcalO(1/k^2)
$$
How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with:
$$
left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)
$$
which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks
integration logarithms asymptotics
asked Jul 22 at 4:07
rubikscube09
869617
add a comment |Â
up vote
1
down vote
favorite
So I have the following integral:
$$
int_k - 1/2^k + 1/2 log t mathrmdt
$$
And I would like to show it can be written as:
$$
int_k - 1/2^k + 1/2 log t mathrmdt = log k + mathcalO(1/k^2)
$$
How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with:
$$
left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)
$$
which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks
integration logarithms asymptotics
asked Jul 22 at 4:07
rubikscube09
869617
1
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So I have the following integral:
$$
int_k - 1/2^k + 1/2 log t mathrmdt
$$
And I would like to show it can be written as:
$$
int_k - 1/2^k + 1/2 log t mathrmdt = log k + mathcalO(1/k^2)
$$
How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with:
$$
left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)
$$
which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks
integration logarithms asymptotics
asked Jul 22 at 4:07
rubikscube09
869617
So I have the following integral:
$$
int_k - 1/2^k + 1/2 log t mathrmdt
$$
And I would like to show it can be written as:
$$
int_k - 1/2^k + 1/2 log t mathrmdt = log k + mathcalO(1/k^2)
$$
How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with:
$$
left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)
$$
which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks
integration logarithms asymptotics
asked Jul 22 at 4:07
rubikscube09
869617
asked Jul 22 at 4:07
rubikscube09
869617
asked Jul 22 at 4:07
rubikscube09
869617
asked Jul 22 at 4:07
rubikscube09
869617
869617
1
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24
add a comment |Â
1
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24
1
1
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $tinleft(k-1/2,k+1/2right)$,
$$
log t=logleft(fractkkright)=log k+logfractk.
$$
Denote
$$
s=fractk-1inleft(-frac12k,frac12kright)subseteqleft(-1,1right).
$$
Thus Taylor's theorem applies, and
$$
log t=log k+logleft(1+sright)=log k+s-fracs^22+O(s^3).
$$
Thanks to this result,
beginalign
int_k-1/2^k+1/2log trm dt&=log k+int_k-1/2^k+1/2left[s-fracs^22+O(s^3)right]rm dt\
&=log k+kint_-1/2k^1/2kleft[s-fracs^22+O(s^3)right]rm ds,
endalign
which suffices to lead to your expected estimate.
answered Jul 22 at 4:43
hypernova
3,489312
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
add a comment |Â
up vote
1
down vote
You properly wrote$$left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)$$
Now rewrite
$$log(x+a)=log(x)+logleft(1+frac a xright)=log(x)+fracax-fraca^22 x^2+fraca^33
x^3+Oleft(frac1x^4right)$$ So,
$$(x+a), (log (x+a)-1)-(x-a),(log (x-a)-1)=2a log(x)-fraca^33 x^2+Oleft(frac1x^4right)$$ and $a=frac 12$.
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $tinleft(k-1/2,k+1/2right)$,
$$
log t=logleft(fractkkright)=log k+logfractk.
$$
Denote
$$
s=fractk-1inleft(-frac12k,frac12kright)subseteqleft(-1,1right).
$$
Thus Taylor's theorem applies, and
$$
log t=log k+logleft(1+sright)=log k+s-fracs^22+O(s^3).
$$
Thanks to this result,
beginalign
int_k-1/2^k+1/2log trm dt&=log k+int_k-1/2^k+1/2left[s-fracs^22+O(s^3)right]rm dt\
&=log k+kint_-1/2k^1/2kleft[s-fracs^22+O(s^3)right]rm ds,
endalign
which suffices to lead to your expected estimate.
answered Jul 22 at 4:43
hypernova
3,489312
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
add a comment |Â
up vote
3
down vote
accepted
For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $tinleft(k-1/2,k+1/2right)$,
$$
log t=logleft(fractkkright)=log k+logfractk.
$$
Denote
$$
s=fractk-1inleft(-frac12k,frac12kright)subseteqleft(-1,1right).
$$
Thus Taylor's theorem applies, and
$$
log t=log k+logleft(1+sright)=log k+s-fracs^22+O(s^3).
$$
Thanks to this result,
beginalign
int_k-1/2^k+1/2log trm dt&=log k+int_k-1/2^k+1/2left[s-fracs^22+O(s^3)right]rm dt\
&=log k+kint_-1/2k^1/2kleft[s-fracs^22+O(s^3)right]rm ds,
endalign
which suffices to lead to your expected estimate.
answered Jul 22 at 4:43
hypernova
3,489312
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $tinleft(k-1/2,k+1/2right)$,
$$
log t=logleft(fractkkright)=log k+logfractk.
$$
Denote
$$
s=fractk-1inleft(-frac12k,frac12kright)subseteqleft(-1,1right).
$$
Thus Taylor's theorem applies, and
$$
log t=log k+logleft(1+sright)=log k+s-fracs^22+O(s^3).
$$
Thanks to this result,
beginalign
int_k-1/2^k+1/2log trm dt&=log k+int_k-1/2^k+1/2left[s-fracs^22+O(s^3)right]rm dt\
&=log k+kint_-1/2k^1/2kleft[s-fracs^22+O(s^3)right]rm ds,
endalign
which suffices to lead to your expected estimate.
answered Jul 22 at 4:43
hypernova
3,489312
For asymptotic analysis, it is not necessary to carry out the exact integral.
Perhaps you may want to play this magic. Note that, for $tinleft(k-1/2,k+1/2right)$,
$$
log t=logleft(fractkkright)=log k+logfractk.
$$
Denote
$$
s=fractk-1inleft(-frac12k,frac12kright)subseteqleft(-1,1right).
$$
Thus Taylor's theorem applies, and
$$
log t=log k+logleft(1+sright)=log k+s-fracs^22+O(s^3).
$$
Thanks to this result,
beginalign
int_k-1/2^k+1/2log trm dt&=log k+int_k-1/2^k+1/2left[s-fracs^22+O(s^3)right]rm dt\
&=log k+kint_-1/2k^1/2kleft[s-fracs^22+O(s^3)right]rm ds,
endalign
which suffices to lead to your expected estimate.
answered Jul 22 at 4:43
hypernova
3,489312
answered Jul 22 at 4:43
hypernova
3,489312
answered Jul 22 at 4:43
hypernova
3,489312
answered Jul 22 at 4:43
hypernova
3,489312
3,489312
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
add a comment |Â
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
Beautiful! Where did you get the intuition to do such a change of variables?
– rubikscube09
Jul 22 at 5:18
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
@rubikscube09: This is because you hope to apply Taylor theorem to $logleft(t/kright)$, while $t/kinleft(1-1/2k,1+1/2kright)$. Thus a natural choice is to expand $logleft(t/kright)$ at $t/k=1$, hence the change of variable follows immediately.
– hypernova
Jul 22 at 9:51
add a comment |Â
up vote
1
down vote
You properly wrote$$left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)$$
Now rewrite
$$log(x+a)=log(x)+logleft(1+frac a xright)=log(x)+fracax-fraca^22 x^2+fraca^33
x^3+Oleft(frac1x^4right)$$ So,
$$(x+a), (log (x+a)-1)-(x-a),(log (x-a)-1)=2a log(x)-fraca^33 x^2+Oleft(frac1x^4right)$$ and $a=frac 12$.
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
You properly wrote$$left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)$$
Now rewrite
$$log(x+a)=log(x)+logleft(1+frac a xright)=log(x)+fracax-fraca^22 x^2+fraca^33
x^3+Oleft(frac1x^4right)$$ So,
$$(x+a), (log (x+a)-1)-(x-a),(log (x-a)-1)=2a log(x)-fraca^33 x^2+Oleft(frac1x^4right)$$ and $a=frac 12$.
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You properly wrote$$left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)$$
Now rewrite
$$log(x+a)=log(x)+logleft(1+frac a xright)=log(x)+fracax-fraca^22 x^2+fraca^33
x^3+Oleft(frac1x^4right)$$ So,
$$(x+a), (log (x+a)-1)-(x-a),(log (x-a)-1)=2a log(x)-fraca^33 x^2+Oleft(frac1x^4right)$$ and $a=frac 12$.
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
You properly wrote$$left(x+frac12right)left(logleft(x+frac12right) - 1right) -left(x-frac12right)left(logleft(x-frac12right) - 1right)$$
Now rewrite
$$log(x+a)=log(x)+logleft(1+frac a xright)=log(x)+fracax-fraca^22 x^2+fraca^33
x^3+Oleft(frac1x^4right)$$ So,
$$(x+a), (log (x+a)-1)-(x-a),(log (x-a)-1)=2a log(x)-fraca^33 x^2+Oleft(frac1x^4right)$$ and $a=frac 12$.
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
answered Jul 22 at 6:21
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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1
Take care that log integral means something different (have a look at en.wikipedia.org/wiki/Logarithmic_integral_function)
– Claude Leibovici
Jul 22 at 6:24