convergence of integral with power and Exponential Function












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try Laplace method.
    $endgroup$
    – Szeto
    Dec 29 '18 at 10:26
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try Laplace method.
    $endgroup$
    – Szeto
    Dec 29 '18 at 10:26














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 10:45







steven

















asked Dec 29 '18 at 9:11









stevensteven

377




377








  • 1




    $begingroup$
    Try Laplace method.
    $endgroup$
    – Szeto
    Dec 29 '18 at 10:26














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your hints
    $endgroup$
    – steven
    Dec 29 '18 at 13:25












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your hints
    $endgroup$
    – steven
    Dec 29 '18 at 13:25
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your hints
    $endgroup$
    – steven
    Dec 29 '18 at 13:25














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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