How to use steepest descent method to approximate $int_0^{1}s^{1/4+i b x}e^{sx}ds$ as $xto+infty$?












2












$begingroup$


Let $0< bleqslant 1$. I am interested in using the steepest descent method to calculate the asymptotic approximation (as $xto+infty$) to the following integral that is related to function ${_1}F_1(1,9/4+ibx,-x)$:



$$J(b,x)=int_0^{1}s^{1/4+i b x}e^{sx}ds=int_0^{1}s^{1/4}exp(x(s+iblog s))dstag{1}$$



Let $rho(z)=z+iblog z$. From several excellent solutions in the posts here and here, I know that we suppose to find contour segments such that $Im(rho(z))$=const. I was not able to find explicit solutions when I set $z=u+iv$ or $z=re^{iphi}$.



Also $rho'(z)=1+ib/z$ so the saddle point is at $z_0=-ib$. And $rho^{"}(z_0)=i/b$.



Thanks for any suggestions.



Update



Set $z=re^{iphi}$ in
$$rho(z)=z+iblog z,$$
we obtain $$Rerho(z)=rcosphi-bphi,quadImrho(z)=rsinphi+blog r$$
To make $Imrho(z)=-c$ we can set
$$sinphi=-frac{blog r}{r}-frac{c}{r},qquadphi=-arcsinleft(frac{blog r}{r}+frac{c}{r}right)$$






complex-analysis asymptotics integral-transforms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
    $endgroup$
    – tired
    Feb 15 '17 at 20:42








  • 1




    $begingroup$
    furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
    $endgroup$
    – tired
    Feb 15 '17 at 20:50








  • 1




    $begingroup$
    to 0 for large z
    $endgroup$
    – tired
    Feb 15 '17 at 20:55










  • $begingroup$
    @tired Thanks for the comments. I will try what you suggested.
    $endgroup$
    – mike
    Feb 16 '17 at 1:08
















2












$begingroup$


Let $0< bleqslant 1$. I am interested in using the steepest descent method to calculate the asymptotic approximation (as $xto+infty$) to the following integral that is related to function ${_1}F_1(1,9/4+ibx,-x)$:



$$J(b,x)=int_0^{1}s^{1/4+i b x}e^{sx}ds=int_0^{1}s^{1/4}exp(x(s+iblog s))dstag{1}$$



Let $rho(z)=z+iblog z$. From several excellent solutions in the posts here and here, I know that we suppose to find contour segments such that $Im(rho(z))$=const. I was not able to find explicit solutions when I set $z=u+iv$ or $z=re^{iphi}$.



Also $rho'(z)=1+ib/z$ so the saddle point is at $z_0=-ib$. And $rho^{"}(z_0)=i/b$.



Thanks for any suggestions.



Update



Set $z=re^{iphi}$ in
$$rho(z)=z+iblog z,$$
we obtain $$Rerho(z)=rcosphi-bphi,quadImrho(z)=rsinphi+blog r$$
To make $Imrho(z)=-c$ we can set
$$sinphi=-frac{blog r}{r}-frac{c}{r},qquadphi=-arcsinleft(frac{blog r}{r}+frac{c}{r}right)$$






complex-analysis asymptotics integral-transforms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
    $endgroup$
    – tired
    Feb 15 '17 at 20:42








  • 1




    $begingroup$
    furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
    $endgroup$
    – tired
    Feb 15 '17 at 20:50








  • 1




    $begingroup$
    to 0 for large z
    $endgroup$
    – tired
    Feb 15 '17 at 20:55










  • $begingroup$
    @tired Thanks for the comments. I will try what you suggested.
    $endgroup$
    – mike
    Feb 16 '17 at 1:08














2












2








2


1



$begingroup$


Let $0< bleqslant 1$. I am interested in using the steepest descent method to calculate the asymptotic approximation (as $xto+infty$) to the following integral that is related to function ${_1}F_1(1,9/4+ibx,-x)$:



$$J(b,x)=int_0^{1}s^{1/4+i b x}e^{sx}ds=int_0^{1}s^{1/4}exp(x(s+iblog s))dstag{1}$$



Let $rho(z)=z+iblog z$. From several excellent solutions in the posts here and here, I know that we suppose to find contour segments such that $Im(rho(z))$=const. I was not able to find explicit solutions when I set $z=u+iv$ or $z=re^{iphi}$.



Also $rho'(z)=1+ib/z$ so the saddle point is at $z_0=-ib$. And $rho^{"}(z_0)=i/b$.



Thanks for any suggestions.



Update



Set $z=re^{iphi}$ in
$$rho(z)=z+iblog z,$$
we obtain $$Rerho(z)=rcosphi-bphi,quadImrho(z)=rsinphi+blog r$$
To make $Imrho(z)=-c$ we can set
$$sinphi=-frac{blog r}{r}-frac{c}{r},qquadphi=-arcsinleft(frac{blog r}{r}+frac{c}{r}right)$$






complex-analysis asymptotics integral-transforms







share|cite|improve this question











$endgroup$




Let $0< bleqslant 1$. I am interested in using the steepest descent method to calculate the asymptotic approximation (as $xto+infty$) to the following integral that is related to function ${_1}F_1(1,9/4+ibx,-x)$:



$$J(b,x)=int_0^{1}s^{1/4+i b x}e^{sx}ds=int_0^{1}s^{1/4}exp(x(s+iblog s))dstag{1}$$



Let $rho(z)=z+iblog z$. From several excellent solutions in the posts here and here, I know that we suppose to find contour segments such that $Im(rho(z))$=const. I was not able to find explicit solutions when I set $z=u+iv$ or $z=re^{iphi}$.



Also $rho'(z)=1+ib/z$ so the saddle point is at $z_0=-ib$. And $rho^{"}(z_0)=i/b$.



Thanks for any suggestions.



Update



Set $z=re^{iphi}$ in
$$rho(z)=z+iblog z,$$
we obtain $$Rerho(z)=rcosphi-bphi,quadImrho(z)=rsinphi+blog r$$
To make $Imrho(z)=-c$ we can set
$$sinphi=-frac{blog r}{r}-frac{c}{r},qquadphi=-arcsinleft(frac{blog r}{r}+frac{c}{r}right)$$






complex-analysis asymptotics integral-transforms




complex-analysis asymptotics integral-transforms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:20









Community♦

1




1










asked Feb 15 '17 at 15:59









mikemike

4,36421019




4,36421019








  • 1




    $begingroup$
    you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
    $endgroup$
    – tired
    Feb 15 '17 at 20:42








  • 1




    $begingroup$
    furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
    $endgroup$
    – tired
    Feb 15 '17 at 20:50








  • 1




    $begingroup$
    to 0 for large z
    $endgroup$
    – tired
    Feb 15 '17 at 20:55










  • $begingroup$
    @tired Thanks for the comments. I will try what you suggested.
    $endgroup$
    – mike
    Feb 16 '17 at 1:08














  • 1




    $begingroup$
    you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
    $endgroup$
    – tired
    Feb 15 '17 at 20:42








  • 1




    $begingroup$
    furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
    $endgroup$
    – tired
    Feb 15 '17 at 20:50








  • 1




    $begingroup$
    to 0 for large z
    $endgroup$
    – tired
    Feb 15 '17 at 20:55










  • $begingroup$
    @tired Thanks for the comments. I will try what you suggested.
    $endgroup$
    – mike
    Feb 16 '17 at 1:08








1




1




$begingroup$
you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
$endgroup$
– tired
Feb 15 '17 at 20:42






$begingroup$
you can expand in a $delta$-neighborhood of the critical points which are the endpoints $z=0,1$ and the saddlepoint at $z=-ib$ so you don't need explicit formulas for the paths of steepest descent
$endgroup$
– tired
Feb 15 '17 at 20:42






1




1




$begingroup$
furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
$endgroup$
– tired
Feb 15 '17 at 20:50






$begingroup$
furthermore, ask yourself the question in which sector of the complex plane $f(z,x)= z^{1/4}e^{xrho(z)}$ converges
$endgroup$
– tired
Feb 15 '17 at 20:50






1




1




$begingroup$
to 0 for large z
$endgroup$
– tired
Feb 15 '17 at 20:55




$begingroup$
to 0 for large z
$endgroup$
– tired
Feb 15 '17 at 20:55












$begingroup$
@tired Thanks for the comments. I will try what you suggested.
$endgroup$
– mike
Feb 16 '17 at 1:08




$begingroup$
@tired Thanks for the comments. I will try what you suggested.
$endgroup$
– mike
Feb 16 '17 at 1:08










1 Answer
1






active

oldest

votes


















0












$begingroup$

If you start with the ${_1hspace{-2px}F_1}$ representation
$$J(b, x) = frac {e^x} {frac 5 4 + i b x}
,{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)},$$
then
$$ lim_{x to infty}
{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)} =
sum_{k = 0}^infty lim_{x to infty}
frac {(-x)^k} {left( frac 9 4 + i b x right)_k} =
sum_{k = 0}^infty left( frac i b right)^k = frac b {b - i}$$
and
$$J(b, x) sim frac {e^x} {(i b + 1) x}.$$
This shows that the asymptotic is not determined by the saddle point at $-i b$.



The application of the steepest descent method would go as follows. The curve $operatorname{Im} rho(z) = text{const}$ passing through $z = 1$ is
$$operatorname{Im} (rho(u + i v) - rho(1)) =
v + frac {b ln(u^2 + v^2)} 2 = 0.$$
The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)xi$, we obtain
$$rho(1 + (1 - i b)xi) = 1 + (1 + b^2) xi + O(xi^2), \
(1 - i b) z^{1/4} e^{x rho(z)} Bigrvert_{z = 1}
int_{-infty}^0 e^{(1 + b^2) x xi} dxi =
frac {e^x} {(i b + 1) x}.$$
It remains to show that other contributions are negligible, using the fact that in the region $operatorname{Re} z < 1 land operatorname{Im} z > 0$, we have
$$operatorname{Re} rho(z) = operatorname{Re} z - b arg z <
operatorname{Re} rho(1).$$






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    0












    $begingroup$

    If you start with the ${_1hspace{-2px}F_1}$ representation
    $$J(b, x) = frac {e^x} {frac 5 4 + i b x}
    ,{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)},$$
    then
    $$ lim_{x to infty}
    {_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)} =
    sum_{k = 0}^infty lim_{x to infty}
    frac {(-x)^k} {left( frac 9 4 + i b x right)_k} =
    sum_{k = 0}^infty left( frac i b right)^k = frac b {b - i}$$
    and
    $$J(b, x) sim frac {e^x} {(i b + 1) x}.$$
    This shows that the asymptotic is not determined by the saddle point at $-i b$.



    The application of the steepest descent method would go as follows. The curve $operatorname{Im} rho(z) = text{const}$ passing through $z = 1$ is
    $$operatorname{Im} (rho(u + i v) - rho(1)) =
    v + frac {b ln(u^2 + v^2)} 2 = 0.$$
    The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)xi$, we obtain
    $$rho(1 + (1 - i b)xi) = 1 + (1 + b^2) xi + O(xi^2), \
    (1 - i b) z^{1/4} e^{x rho(z)} Bigrvert_{z = 1}
    int_{-infty}^0 e^{(1 + b^2) x xi} dxi =
    frac {e^x} {(i b + 1) x}.$$
    It remains to show that other contributions are negligible, using the fact that in the region $operatorname{Re} z < 1 land operatorname{Im} z > 0$, we have
    $$operatorname{Re} rho(z) = operatorname{Re} z - b arg z <
    operatorname{Re} rho(1).$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If you start with the ${_1hspace{-2px}F_1}$ representation
      $$J(b, x) = frac {e^x} {frac 5 4 + i b x}
      ,{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)},$$
      then
      $$ lim_{x to infty}
      {_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)} =
      sum_{k = 0}^infty lim_{x to infty}
      frac {(-x)^k} {left( frac 9 4 + i b x right)_k} =
      sum_{k = 0}^infty left( frac i b right)^k = frac b {b - i}$$
      and
      $$J(b, x) sim frac {e^x} {(i b + 1) x}.$$
      This shows that the asymptotic is not determined by the saddle point at $-i b$.



      The application of the steepest descent method would go as follows. The curve $operatorname{Im} rho(z) = text{const}$ passing through $z = 1$ is
      $$operatorname{Im} (rho(u + i v) - rho(1)) =
      v + frac {b ln(u^2 + v^2)} 2 = 0.$$
      The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)xi$, we obtain
      $$rho(1 + (1 - i b)xi) = 1 + (1 + b^2) xi + O(xi^2), \
      (1 - i b) z^{1/4} e^{x rho(z)} Bigrvert_{z = 1}
      int_{-infty}^0 e^{(1 + b^2) x xi} dxi =
      frac {e^x} {(i b + 1) x}.$$
      It remains to show that other contributions are negligible, using the fact that in the region $operatorname{Re} z < 1 land operatorname{Im} z > 0$, we have
      $$operatorname{Re} rho(z) = operatorname{Re} z - b arg z <
      operatorname{Re} rho(1).$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If you start with the ${_1hspace{-2px}F_1}$ representation
        $$J(b, x) = frac {e^x} {frac 5 4 + i b x}
        ,{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)},$$
        then
        $$ lim_{x to infty}
        {_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)} =
        sum_{k = 0}^infty lim_{x to infty}
        frac {(-x)^k} {left( frac 9 4 + i b x right)_k} =
        sum_{k = 0}^infty left( frac i b right)^k = frac b {b - i}$$
        and
        $$J(b, x) sim frac {e^x} {(i b + 1) x}.$$
        This shows that the asymptotic is not determined by the saddle point at $-i b$.



        The application of the steepest descent method would go as follows. The curve $operatorname{Im} rho(z) = text{const}$ passing through $z = 1$ is
        $$operatorname{Im} (rho(u + i v) - rho(1)) =
        v + frac {b ln(u^2 + v^2)} 2 = 0.$$
        The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)xi$, we obtain
        $$rho(1 + (1 - i b)xi) = 1 + (1 + b^2) xi + O(xi^2), \
        (1 - i b) z^{1/4} e^{x rho(z)} Bigrvert_{z = 1}
        int_{-infty}^0 e^{(1 + b^2) x xi} dxi =
        frac {e^x} {(i b + 1) x}.$$
        It remains to show that other contributions are negligible, using the fact that in the region $operatorname{Re} z < 1 land operatorname{Im} z > 0$, we have
        $$operatorname{Re} rho(z) = operatorname{Re} z - b arg z <
        operatorname{Re} rho(1).$$






        share|cite|improve this answer











        $endgroup$



        If you start with the ${_1hspace{-2px}F_1}$ representation
        $$J(b, x) = frac {e^x} {frac 5 4 + i b x}
        ,{_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)},$$
        then
        $$ lim_{x to infty}
        {_1hspace{-2px}F_1} {left( 1, frac 9 4 + i b x, -x right)} =
        sum_{k = 0}^infty lim_{x to infty}
        frac {(-x)^k} {left( frac 9 4 + i b x right)_k} =
        sum_{k = 0}^infty left( frac i b right)^k = frac b {b - i}$$
        and
        $$J(b, x) sim frac {e^x} {(i b + 1) x}.$$
        This shows that the asymptotic is not determined by the saddle point at $-i b$.



        The application of the steepest descent method would go as follows. The curve $operatorname{Im} rho(z) = text{const}$ passing through $z = 1$ is
        $$operatorname{Im} (rho(u + i v) - rho(1)) =
        v + frac {b ln(u^2 + v^2)} 2 = 0.$$
        The slope at $z = 1$ is $-b$. Taking $z = 1 + (1 - i b)xi$, we obtain
        $$rho(1 + (1 - i b)xi) = 1 + (1 + b^2) xi + O(xi^2), \
        (1 - i b) z^{1/4} e^{x rho(z)} Bigrvert_{z = 1}
        int_{-infty}^0 e^{(1 + b^2) x xi} dxi =
        frac {e^x} {(i b + 1) x}.$$
        It remains to show that other contributions are negligible, using the fact that in the region $operatorname{Re} z < 1 land operatorname{Im} z > 0$, we have
        $$operatorname{Re} rho(z) = operatorname{Re} z - b arg z <
        operatorname{Re} rho(1).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 2:51

























        answered Apr 23 '18 at 19:20









        MaximMaxim

        4,7281219




        4,7281219






























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