Compute $int_0^infty e^{-az} sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz$












2












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I got stuck in this problem:



$$int_0^infty e^{-az} sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz ~~(1)$$
where $a>0$.



My thoughts: By Binomial expansion we have
$$ int_0^infty e^{-az} z^{n+2}(u+z)^ndz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} int_0^infty z^{2n-k+2}e^{-az}dz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.~~~~~~~~
$$
So
$$(1)=sum_{n=0}^infty frac{1}{(n+1)!(n+2)!} sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.
$$
Then I got lost. Please let me know if you have any idea or direction about computing $(1)$. Any help would be greatly appreciated, thanks!






calculus combinatorics bessel-functions







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    It should have a simplified solution but I really don't know how to proceed.
    $endgroup$
    – Violet. W
    Feb 6 at 17:28
















2












$begingroup$


I got stuck in this problem:



$$int_0^infty e^{-az} sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz ~~(1)$$
where $a>0$.



My thoughts: By Binomial expansion we have
$$ int_0^infty e^{-az} z^{n+2}(u+z)^ndz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} int_0^infty z^{2n-k+2}e^{-az}dz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.~~~~~~~~
$$
So
$$(1)=sum_{n=0}^infty frac{1}{(n+1)!(n+2)!} sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.
$$
Then I got lost. Please let me know if you have any idea or direction about computing $(1)$. Any help would be greatly appreciated, thanks!






calculus combinatorics bessel-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    It should have a simplified solution but I really don't know how to proceed.
    $endgroup$
    – Violet. W
    Feb 6 at 17:28














2












2








2


1



$begingroup$


I got stuck in this problem:



$$int_0^infty e^{-az} sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz ~~(1)$$
where $a>0$.



My thoughts: By Binomial expansion we have
$$ int_0^infty e^{-az} z^{n+2}(u+z)^ndz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} int_0^infty z^{2n-k+2}e^{-az}dz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.~~~~~~~~
$$
So
$$(1)=sum_{n=0}^infty frac{1}{(n+1)!(n+2)!} sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.
$$
Then I got lost. Please let me know if you have any idea or direction about computing $(1)$. Any help would be greatly appreciated, thanks!






calculus combinatorics bessel-functions







share|cite|improve this question









$endgroup$




I got stuck in this problem:



$$int_0^infty e^{-az} sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} dz ~~(1)$$
where $a>0$.



My thoughts: By Binomial expansion we have
$$ int_0^infty e^{-az} z^{n+2}(u+z)^ndz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} int_0^infty z^{2n-k+2}e^{-az}dz\
= sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.~~~~~~~~
$$
So
$$(1)=sum_{n=0}^infty frac{1}{(n+1)!(n+2)!} sum_{k=0}^n frac{n!u^k}{k!(n-k)!} frac{(2n-k+2)!}{a^{2n-k+3}}.
$$
Then I got lost. Please let me know if you have any idea or direction about computing $(1)$. Any help would be greatly appreciated, thanks!






calculus combinatorics bessel-functions




calculus combinatorics bessel-functions






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asked Feb 2 at 2:53









Violet. WViolet. W

363




363












  • $begingroup$
    It should have a simplified solution but I really don't know how to proceed.
    $endgroup$
    – Violet. W
    Feb 6 at 17:28


















  • $begingroup$
    It should have a simplified solution but I really don't know how to proceed.
    $endgroup$
    – Violet. W
    Feb 6 at 17:28
















$begingroup$
It should have a simplified solution but I really don't know how to proceed.
$endgroup$
– Violet. W
Feb 6 at 17:28




$begingroup$
It should have a simplified solution but I really don't know how to proceed.
$endgroup$
– Violet. W
Feb 6 at 17:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

An explicit form for the series can be found. From the series expansion for the modified Bessel function
begin{equation}
I_{1}left(zright)=sum_{k=0}^{infty}frac{(tfrac{z}{2})^{2k+1}}{k!left(k+1right)!}
end{equation}
we have
begin{align}
sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{n=0}^infty frac{left[ z(z+u) right]^{n+3/2}}{(n+1)!(n+2)!}\
&=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ z(z+u) right]^{k+1/2}}{k!(k+1)!}\
&=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ sqrt{z(z+u) }right]^{2k+1}}{k!(k+1)!}\
&=frac{sqrt{z}}{left( u+z right)^{3/2}}left[I_1left( 2sqrt{z(z+u) } right)-sqrt{z(z+u) }right]\
&=frac{sqrt{z}}{left( u+z right)^{3/2}}I_1left( 2sqrt{z(z+u) } right)-frac{z}{z+u}
end{align}
To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
begin{align}
int_0^infty &t^{mu-1}left( t+beta right)^{-mu}I_{2nu}left( alphaleft( t^2+beta t right)^{1/2} right)e^{-pt},dt\
&=frac{2Gammaleft( mu+nu right)e^{beta p/2}}{alphabetaGamma(2nu+1)}M_{1/2-mu,nu}left( frac{alpha^2beta}{2p+sqrt{p^2-alpha^2}} right)W_{1/2-mu,nu}left( frac{betaleft( p+sqrt{p^2-alpha^2} right)}{2} right)
end{align}
(with $mu=3/2, nu=1/2,beta=u,alpha=2$) and the integral representation of the exponential integral (DLMF)
begin{align}
int_0^inftyfrac{ze^{-az}}{z+u},dz&=int_0^infty e^{-az},dz-uint_0^inftyfrac{e^{-az}}{z+u},dz\
&=frac{1}{a}-ue^{au}E_{1}left( au right)
end{align}






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    1 Answer
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    1 Answer
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    active

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    active

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    1












    $begingroup$

    An explicit form for the series can be found. From the series expansion for the modified Bessel function
    begin{equation}
    I_{1}left(zright)=sum_{k=0}^{infty}frac{(tfrac{z}{2})^{2k+1}}{k!left(k+1right)!}
    end{equation}
    we have
    begin{align}
    sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{n=0}^infty frac{left[ z(z+u) right]^{n+3/2}}{(n+1)!(n+2)!}\
    &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ z(z+u) right]^{k+1/2}}{k!(k+1)!}\
    &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ sqrt{z(z+u) }right]^{2k+1}}{k!(k+1)!}\
    &=frac{sqrt{z}}{left( u+z right)^{3/2}}left[I_1left( 2sqrt{z(z+u) } right)-sqrt{z(z+u) }right]\
    &=frac{sqrt{z}}{left( u+z right)^{3/2}}I_1left( 2sqrt{z(z+u) } right)-frac{z}{z+u}
    end{align}
    To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
    begin{align}
    int_0^infty &t^{mu-1}left( t+beta right)^{-mu}I_{2nu}left( alphaleft( t^2+beta t right)^{1/2} right)e^{-pt},dt\
    &=frac{2Gammaleft( mu+nu right)e^{beta p/2}}{alphabetaGamma(2nu+1)}M_{1/2-mu,nu}left( frac{alpha^2beta}{2p+sqrt{p^2-alpha^2}} right)W_{1/2-mu,nu}left( frac{betaleft( p+sqrt{p^2-alpha^2} right)}{2} right)
    end{align}
    (with $mu=3/2, nu=1/2,beta=u,alpha=2$) and the integral representation of the exponential integral (DLMF)
    begin{align}
    int_0^inftyfrac{ze^{-az}}{z+u},dz&=int_0^infty e^{-az},dz-uint_0^inftyfrac{e^{-az}}{z+u},dz\
    &=frac{1}{a}-ue^{au}E_{1}left( au right)
    end{align}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      An explicit form for the series can be found. From the series expansion for the modified Bessel function
      begin{equation}
      I_{1}left(zright)=sum_{k=0}^{infty}frac{(tfrac{z}{2})^{2k+1}}{k!left(k+1right)!}
      end{equation}
      we have
      begin{align}
      sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{n=0}^infty frac{left[ z(z+u) right]^{n+3/2}}{(n+1)!(n+2)!}\
      &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ z(z+u) right]^{k+1/2}}{k!(k+1)!}\
      &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ sqrt{z(z+u) }right]^{2k+1}}{k!(k+1)!}\
      &=frac{sqrt{z}}{left( u+z right)^{3/2}}left[I_1left( 2sqrt{z(z+u) } right)-sqrt{z(z+u) }right]\
      &=frac{sqrt{z}}{left( u+z right)^{3/2}}I_1left( 2sqrt{z(z+u) } right)-frac{z}{z+u}
      end{align}
      To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
      begin{align}
      int_0^infty &t^{mu-1}left( t+beta right)^{-mu}I_{2nu}left( alphaleft( t^2+beta t right)^{1/2} right)e^{-pt},dt\
      &=frac{2Gammaleft( mu+nu right)e^{beta p/2}}{alphabetaGamma(2nu+1)}M_{1/2-mu,nu}left( frac{alpha^2beta}{2p+sqrt{p^2-alpha^2}} right)W_{1/2-mu,nu}left( frac{betaleft( p+sqrt{p^2-alpha^2} right)}{2} right)
      end{align}
      (with $mu=3/2, nu=1/2,beta=u,alpha=2$) and the integral representation of the exponential integral (DLMF)
      begin{align}
      int_0^inftyfrac{ze^{-az}}{z+u},dz&=int_0^infty e^{-az},dz-uint_0^inftyfrac{e^{-az}}{z+u},dz\
      &=frac{1}{a}-ue^{au}E_{1}left( au right)
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        An explicit form for the series can be found. From the series expansion for the modified Bessel function
        begin{equation}
        I_{1}left(zright)=sum_{k=0}^{infty}frac{(tfrac{z}{2})^{2k+1}}{k!left(k+1right)!}
        end{equation}
        we have
        begin{align}
        sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{n=0}^infty frac{left[ z(z+u) right]^{n+3/2}}{(n+1)!(n+2)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ z(z+u) right]^{k+1/2}}{k!(k+1)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ sqrt{z(z+u) }right]^{2k+1}}{k!(k+1)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}left[I_1left( 2sqrt{z(z+u) } right)-sqrt{z(z+u) }right]\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}I_1left( 2sqrt{z(z+u) } right)-frac{z}{z+u}
        end{align}
        To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
        begin{align}
        int_0^infty &t^{mu-1}left( t+beta right)^{-mu}I_{2nu}left( alphaleft( t^2+beta t right)^{1/2} right)e^{-pt},dt\
        &=frac{2Gammaleft( mu+nu right)e^{beta p/2}}{alphabetaGamma(2nu+1)}M_{1/2-mu,nu}left( frac{alpha^2beta}{2p+sqrt{p^2-alpha^2}} right)W_{1/2-mu,nu}left( frac{betaleft( p+sqrt{p^2-alpha^2} right)}{2} right)
        end{align}
        (with $mu=3/2, nu=1/2,beta=u,alpha=2$) and the integral representation of the exponential integral (DLMF)
        begin{align}
        int_0^inftyfrac{ze^{-az}}{z+u},dz&=int_0^infty e^{-az},dz-uint_0^inftyfrac{e^{-az}}{z+u},dz\
        &=frac{1}{a}-ue^{au}E_{1}left( au right)
        end{align}






        share|cite|improve this answer









        $endgroup$



        An explicit form for the series can be found. From the series expansion for the modified Bessel function
        begin{equation}
        I_{1}left(zright)=sum_{k=0}^{infty}frac{(tfrac{z}{2})^{2k+1}}{k!left(k+1right)!}
        end{equation}
        we have
        begin{align}
        sum_{n=0}^infty frac{z^{n+2}(u+z)^n}{(n+1)!(n+2)!} &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{n=0}^infty frac{left[ z(z+u) right]^{n+3/2}}{(n+1)!(n+2)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ z(z+u) right]^{k+1/2}}{k!(k+1)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}sum_{k=1}^infty frac{left[ sqrt{z(z+u) }right]^{2k+1}}{k!(k+1)!}\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}left[I_1left( 2sqrt{z(z+u) } right)-sqrt{z(z+u) }right]\
        &=frac{sqrt{z}}{left( u+z right)^{3/2}}I_1left( 2sqrt{z(z+u) } right)-frac{z}{z+u}
        end{align}
        To calculate the Laplace transform, we can use an identity tabulated in Ederlyi TI 4.17.15:
        begin{align}
        int_0^infty &t^{mu-1}left( t+beta right)^{-mu}I_{2nu}left( alphaleft( t^2+beta t right)^{1/2} right)e^{-pt},dt\
        &=frac{2Gammaleft( mu+nu right)e^{beta p/2}}{alphabetaGamma(2nu+1)}M_{1/2-mu,nu}left( frac{alpha^2beta}{2p+sqrt{p^2-alpha^2}} right)W_{1/2-mu,nu}left( frac{betaleft( p+sqrt{p^2-alpha^2} right)}{2} right)
        end{align}
        (with $mu=3/2, nu=1/2,beta=u,alpha=2$) and the integral representation of the exponential integral (DLMF)
        begin{align}
        int_0^inftyfrac{ze^{-az}}{z+u},dz&=int_0^infty e^{-az},dz-uint_0^inftyfrac{e^{-az}}{z+u},dz\
        &=frac{1}{a}-ue^{au}E_{1}left( au right)
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Mar 31 at 21:10









        Paul EntaPaul Enta

        5,47611435




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