convergence of integral with power and Exponential Function
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Please prove that: for $xin(0,1/2)$ and $a<0$,
$int_{0}^{1}v^{-x}(1-v)^{-x}e^{a(1-v)M}dv$ tends to zero as $M$ goes to infinite.
Many thanks for your help!
Maybe we can use speicial functions
$int _ { 0 } ^ { u } x ^ { nu - 1 } ( u - x ) ^ { mu - 1 } e ^ { beta x } d x = mathrm { B } ( mu , nu ) u ^ { mu + nu - 1 } quad 1 F _ { 1 } ( nu ; mu + nu ; beta u )$
How to use this formula!
Many thanks for your help!
calculus integration inequality
asked Dec 29 '18 at 9:11
stevensteven
377
$endgroup$
add a comment |
$begingroup$
Please prove that: for $xin(0,1/2)$ and $a<0$,
$int_{0}^{1}v^{-x}(1-v)^{-x}e^{a(1-v)M}dv$ tends to zero as $M$ goes to infinite.
Many thanks for your help!
Maybe we can use speicial functions
$int _ { 0 } ^ { u } x ^ { nu - 1 } ( u - x ) ^ { mu - 1 } e ^ { beta x } d x = mathrm { B } ( mu , nu ) u ^ { mu + nu - 1 } quad 1 F _ { 1 } ( nu ; mu + nu ; beta u )$
How to use this formula!
Many thanks for your help!
calculus integration inequality
asked Dec 29 '18 at 9:11
stevensteven
377
$endgroup$
1
$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26
add a comment |
$begingroup$
Please prove that: for $xin(0,1/2)$ and $a<0$,
$int_{0}^{1}v^{-x}(1-v)^{-x}e^{a(1-v)M}dv$ tends to zero as $M$ goes to infinite.
Many thanks for your help!
Maybe we can use speicial functions
$int _ { 0 } ^ { u } x ^ { nu - 1 } ( u - x ) ^ { mu - 1 } e ^ { beta x } d x = mathrm { B } ( mu , nu ) u ^ { mu + nu - 1 } quad 1 F _ { 1 } ( nu ; mu + nu ; beta u )$
How to use this formula!
Many thanks for your help!
calculus integration inequality
asked Dec 29 '18 at 9:11
stevensteven
377
$endgroup$
Please prove that: for $xin(0,1/2)$ and $a<0$,
$int_{0}^{1}v^{-x}(1-v)^{-x}e^{a(1-v)M}dv$ tends to zero as $M$ goes to infinite.
Many thanks for your help!
Maybe we can use speicial functions
$int _ { 0 } ^ { u } x ^ { nu - 1 } ( u - x ) ^ { mu - 1 } e ^ { beta x } d x = mathrm { B } ( mu , nu ) u ^ { mu + nu - 1 } quad 1 F _ { 1 } ( nu ; mu + nu ; beta u )$
How to use this formula!
Many thanks for your help!
calculus integration inequality
calculus integration inequality
asked Dec 29 '18 at 9:11
stevensteven
377
asked Dec 29 '18 at 9:11
stevensteven
377
edited Dec 29 '18 at 10:45
steven
asked Dec 29 '18 at 9:11
stevensteven
377
asked Dec 29 '18 at 9:11
stevensteven
377
asked Dec 29 '18 at 9:11
stevensteven
377
377
1
$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26
add a comment |
1
$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26
1
1
$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26
$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $Mto infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $Mto infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
add a comment |
$begingroup$
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $Mto infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
add a comment |
$begingroup$
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $Mto infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
$endgroup$
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $Mto infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
answered Dec 29 '18 at 12:41
Kavi Rama MurthyKavi Rama Murthy
76.1k53370
76.1k53370
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
add a comment |
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
$begingroup$
Thanks for your hints
$endgroup$
– steven
Dec 29 '18 at 13:25
add a comment |
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$begingroup$
Try Laplace method.
$endgroup$
– Szeto
Dec 29 '18 at 10:26