Finite integral and limit
up vote
1
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Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 at 22:59
ZAF
3706
add a comment |
up vote
1
down vote
favorite
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 at 22:59
ZAF
3706
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 at 22:59
ZAF
3706
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
limits improper-integrals
asked Nov 21 at 22:59
ZAF
3706
asked Nov 21 at 22:59
ZAF
3706
edited Nov 21 at 23:13
asked Nov 21 at 22:59
ZAF
3706
asked Nov 21 at 22:59
ZAF
3706
asked Nov 21 at 22:59
ZAF
3706
3706
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14
add a comment |
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
add a comment |
up vote
2
down vote
accepted
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
answered Nov 21 at 23:36
Kavi Rama Murthy
41.8k31751
41.8k31751
add a comment |
add a comment |
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I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 at 23:14