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$begingroup$
I'm trying to calculate this limit:
$$lim_{xtoinfty} left(int_0^1 t^{-tx} dtright)^{frac1x}$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate this limit:
$$lim_{xtoinfty} left(int_0^1 t^{-tx} dtright)^{frac1x}$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
2
$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
$begingroup$
I'm trying to calculate this limit:
$$lim_{xtoinfty} left(int_0^1 t^{-tx} dtright)^{frac1x}$$
I tried the squeezing idea without success.
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
$endgroup$
I'm trying to calculate this limit:
$$lim_{xtoinfty} left(int_0^1 t^{-tx} dtright)^{frac1x}$$
I tried the squeezing idea without success.
calculus limits definite-integrals
calculus limits definite-integrals
edited Mar 17 at 23:35
RRL
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
edited Mar 17 at 23:35
RRL
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
edited Mar 17 at 23:35
RRL
53.2k52574
edited Mar 17 at 23:35
RRL
53.2k52574
edited Mar 17 at 23:35
RRL
53.2k52574
53.2k52574
asked Feb 23 '17 at 19:13
song01song01
613
asked Feb 23 '17 at 19:13
song01song01
613
asked Feb 23 '17 at 19:13
song01song01
613
613
2
$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
2
$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
2
2
$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any large $x$ we have
$$ int_{0}^{1}t^{-tx},dt = int_{0}^{1}expleft(-x tlog tright),dt = sum_{ngeq 0}frac{x^n}{n!}int_{0}^{1}t^n(-log t)^n,dt =sum_{ngeq 0}frac{x^n}{n!(n+1)^{n+1}}$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_{xto +infty}frac{log f(x)}{x} stackrel{dH}{=} explim_{xto +infty}frac{f'(x)}{f(x)}=explim_{xto +infty}frac{int_{0}^{1}(-tlog t)t^{-tx},dt}{int_{0}^{1}t^{-tx},dt}$$
that is $color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $Ccdotdeltaleft(t-frac{1}{e}right)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$
With $0 < t < 1,$ we have
$$t^{-tx} = exp(-t ln t)^x leqslant exp(e^{-1})^x \ implies left(int_0^1 t^{-tx} , dtright)^{1/x} leqslant exp(e^{-1}).$$
Using continuity, for any small $epsilon$, there is an interval $[e^{-1} - delta, e^{-1} + delta] subset [0,1]$ where $t^{-t} > exp(e^{-1}) - epsilon$ and
$$left(int_0^1 t^{-tx} , dtright)^{1/x} geqslant left(int_{exp(e^{-1}) - delta}^{exp(e^{-1}) + delta} t^{-tx} , dtright)^{1/x} > (exp(e^{-1})- epsilon)(2 delta)^{1/x}. $$
Since $(2delta)^{1/x} to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_{x to infty}left(int_0^1 t^{-tx} , dtright)^{1/x} = exp(e^{-1})$$
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
For any large $x$ we have
$$ int_{0}^{1}t^{-tx},dt = int_{0}^{1}expleft(-x tlog tright),dt = sum_{ngeq 0}frac{x^n}{n!}int_{0}^{1}t^n(-log t)^n,dt =sum_{ngeq 0}frac{x^n}{n!(n+1)^{n+1}}$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_{xto +infty}frac{log f(x)}{x} stackrel{dH}{=} explim_{xto +infty}frac{f'(x)}{f(x)}=explim_{xto +infty}frac{int_{0}^{1}(-tlog t)t^{-tx},dt}{int_{0}^{1}t^{-tx},dt}$$
that is $color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $Ccdotdeltaleft(t-frac{1}{e}right)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
For any large $x$ we have
$$ int_{0}^{1}t^{-tx},dt = int_{0}^{1}expleft(-x tlog tright),dt = sum_{ngeq 0}frac{x^n}{n!}int_{0}^{1}t^n(-log t)^n,dt =sum_{ngeq 0}frac{x^n}{n!(n+1)^{n+1}}$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_{xto +infty}frac{log f(x)}{x} stackrel{dH}{=} explim_{xto +infty}frac{f'(x)}{f(x)}=explim_{xto +infty}frac{int_{0}^{1}(-tlog t)t^{-tx},dt}{int_{0}^{1}t^{-tx},dt}$$
that is $color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $Ccdotdeltaleft(t-frac{1}{e}right)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
For any large $x$ we have
$$ int_{0}^{1}t^{-tx},dt = int_{0}^{1}expleft(-x tlog tright),dt = sum_{ngeq 0}frac{x^n}{n!}int_{0}^{1}t^n(-log t)^n,dt =sum_{ngeq 0}frac{x^n}{n!(n+1)^{n+1}}$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_{xto +infty}frac{log f(x)}{x} stackrel{dH}{=} explim_{xto +infty}frac{f'(x)}{f(x)}=explim_{xto +infty}frac{int_{0}^{1}(-tlog t)t^{-tx},dt}{int_{0}^{1}t^{-tx},dt}$$
that is $color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $Ccdotdeltaleft(t-frac{1}{e}right)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
For any large $x$ we have
$$ int_{0}^{1}t^{-tx},dt = int_{0}^{1}expleft(-x tlog tright),dt = sum_{ngeq 0}frac{x^n}{n!}int_{0}^{1}t^n(-log t)^n,dt =sum_{ngeq 0}frac{x^n}{n!(n+1)^{n+1}}$$
If we call this entire function $f(x)$, the wanted limit equals
$$explim_{xto +infty}frac{log f(x)}{x} stackrel{dH}{=} explim_{xto +infty}frac{f'(x)}{f(x)}=explim_{xto +infty}frac{int_{0}^{1}(-tlog t)t^{-tx},dt}{int_{0}^{1}t^{-tx},dt}$$
that is $color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $Ccdotdeltaleft(t-frac{1}{e}right)$.
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 23 '17 at 19:29
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
add a comment |
add a comment |
$begingroup$
A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$
With $0 < t < 1,$ we have
$$t^{-tx} = exp(-t ln t)^x leqslant exp(e^{-1})^x \ implies left(int_0^1 t^{-tx} , dtright)^{1/x} leqslant exp(e^{-1}).$$
Using continuity, for any small $epsilon$, there is an interval $[e^{-1} - delta, e^{-1} + delta] subset [0,1]$ where $t^{-t} > exp(e^{-1}) - epsilon$ and
$$left(int_0^1 t^{-tx} , dtright)^{1/x} geqslant left(int_{exp(e^{-1}) - delta}^{exp(e^{-1}) + delta} t^{-tx} , dtright)^{1/x} > (exp(e^{-1})- epsilon)(2 delta)^{1/x}. $$
Since $(2delta)^{1/x} to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_{x to infty}left(int_0^1 t^{-tx} , dtright)^{1/x} = exp(e^{-1})$$
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$
With $0 < t < 1,$ we have
$$t^{-tx} = exp(-t ln t)^x leqslant exp(e^{-1})^x \ implies left(int_0^1 t^{-tx} , dtright)^{1/x} leqslant exp(e^{-1}).$$
Using continuity, for any small $epsilon$, there is an interval $[e^{-1} - delta, e^{-1} + delta] subset [0,1]$ where $t^{-t} > exp(e^{-1}) - epsilon$ and
$$left(int_0^1 t^{-tx} , dtright)^{1/x} geqslant left(int_{exp(e^{-1}) - delta}^{exp(e^{-1}) + delta} t^{-tx} , dtright)^{1/x} > (exp(e^{-1})- epsilon)(2 delta)^{1/x}. $$
Since $(2delta)^{1/x} to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_{x to infty}left(int_0^1 t^{-tx} , dtright)^{1/x} = exp(e^{-1})$$
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
$endgroup$
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$
With $0 < t < 1,$ we have
$$t^{-tx} = exp(-t ln t)^x leqslant exp(e^{-1})^x \ implies left(int_0^1 t^{-tx} , dtright)^{1/x} leqslant exp(e^{-1}).$$
Using continuity, for any small $epsilon$, there is an interval $[e^{-1} - delta, e^{-1} + delta] subset [0,1]$ where $t^{-t} > exp(e^{-1}) - epsilon$ and
$$left(int_0^1 t^{-tx} , dtright)^{1/x} geqslant left(int_{exp(e^{-1}) - delta}^{exp(e^{-1}) + delta} t^{-tx} , dtright)^{1/x} > (exp(e^{-1})- epsilon)(2 delta)^{1/x}. $$
Since $(2delta)^{1/x} to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_{x to infty}left(int_0^1 t^{-tx} , dtright)^{1/x} = exp(e^{-1})$$
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
$endgroup$
A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$
With $0 < t < 1,$ we have
$$t^{-tx} = exp(-t ln t)^x leqslant exp(e^{-1})^x \ implies left(int_0^1 t^{-tx} , dtright)^{1/x} leqslant exp(e^{-1}).$$
Using continuity, for any small $epsilon$, there is an interval $[e^{-1} - delta, e^{-1} + delta] subset [0,1]$ where $t^{-t} > exp(e^{-1}) - epsilon$ and
$$left(int_0^1 t^{-tx} , dtright)^{1/x} geqslant left(int_{exp(e^{-1}) - delta}^{exp(e^{-1}) + delta} t^{-tx} , dtright)^{1/x} > (exp(e^{-1})- epsilon)(2 delta)^{1/x}. $$
Since $(2delta)^{1/x} to 1$ as $x to infty$ and $epsilon$ can be arbitrarily small, the squeeze theorem yields
$$lim_{x to infty}left(int_0^1 t^{-tx} , dtright)^{1/x} = exp(e^{-1})$$
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
edited Feb 23 '17 at 19:46
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
answered Feb 23 '17 at 19:25
RRLRRL
53.2k52574
53.2k52574
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
$begingroup$
Perfect use of squeeze +1
$endgroup$
– Paramanand Singh
Feb 24 '17 at 6:38
add a comment |
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$begingroup$
If $f$ is continuous or just integrable , $lim_{n to infty} left(int_0^1 |f(x)|^n, dxright)^{1/n} = sup_{x in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted.
$endgroup$
– RRL
Feb 23 '17 at 19:30