New bounds for $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}$
I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
in the form of this approximation :
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$
now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.
The question is how do I 'tighten' up the bounds on this particular approximation (2)?
What are the strategies for going about this sort of thing?
integration definite-integrals asymptotics approximation
asked Jan 4 at 14:08
onepoundonepound
331216
add a comment |
I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
in the form of this approximation :
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$
now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.
The question is how do I 'tighten' up the bounds on this particular approximation (2)?
What are the strategies for going about this sort of thing?
integration definite-integrals asymptotics approximation
asked Jan 4 at 14:08
onepoundonepound
331216
add a comment |
I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
in the form of this approximation :
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$
now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.
The question is how do I 'tighten' up the bounds on this particular approximation (2)?
What are the strategies for going about this sort of thing?
integration definite-integrals asymptotics approximation
asked Jan 4 at 14:08
onepoundonepound
331216
I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
in the form of this approximation :
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$
now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.
The question is how do I 'tighten' up the bounds on this particular approximation (2)?
What are the strategies for going about this sort of thing?
integration definite-integrals asymptotics approximation
integration definite-integrals asymptotics approximation
asked Jan 4 at 14:08
onepoundonepound
331216
asked Jan 4 at 14:08
onepoundonepound
331216
edited Jan 4 at 14:27
onepound
asked Jan 4 at 14:08
onepoundonepound
331216
asked Jan 4 at 14:08
onepoundonepound
331216
asked Jan 4 at 14:08
onepoundonepound
331216
331216
add a comment |
add a comment |
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To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.
Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$
answered Jan 4 at 14:09
onepoundonepound
331216
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.
Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$
answered Jan 4 at 14:09
onepoundonepound
331216
add a comment |
To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.
Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$
answered Jan 4 at 14:09
onepoundonepound
331216
add a comment |
To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.
Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$
answered Jan 4 at 14:09
onepoundonepound
331216
To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.
Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:
$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$
answered Jan 4 at 14:09
onepoundonepound
331216
answered Jan 4 at 14:09
onepoundonepound
331216
answered Jan 4 at 14:09
onepoundonepound
331216
answered Jan 4 at 14:09
onepoundonepound
331216
331216
add a comment |
add a comment |
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