Laplace transform of “shifted” modified Bessel function












0














Dear all: i'm trying to derive a closed form the following integral,
$$
X_n(R)=int_0^infty exp(-p, r)I_n(omega (r+R)), dr,
$$

where $I_n$ is the standard modified Bessel function of the first kind and $p>omega$, $0 leq omega<1$. I know that, when there is no shift ($R=0$), the values $X_n(R=0)$ are well-known because they are standard Laplace transforms $L[I_n(omega, cdot)](p)$, see for instance here.



Yet, for $R not =0$, i've come up with 2 possibilities,



-- using a "shift theorem" for the Laplace transform, which means that i change variable $r to r+R$ in order to get an exponential term $exp(pR)$ outside. However, this has an effect on the interval of integration,
$$
X_n(R)=exp(pR) int_R^infty exp(-pr)I_n(omega, r), dr,
$$

so that the value of the integral isn't well-known anymore. If $|R| ll 1$, one may use a polynomial approximation of $I_n$ in order to (approximately) compute the term
$$
Y_n(R)=int_0^R exp(-pr)I_n(omega, r), dr simeq int_0^R exp(-pr)frac{x^n}{2^n ,n!} , dr,
$$

which should be subtracted from the value of the Laplace transform as follows:
$$
X_n(R)=exp(pR)Big(L[I_n(omega, cdot)](p)-Y_n(R) Big).
$$



-- using the "summation formula" for Bessel functions,
$$
I_n(a+b)=sum_{k=-infty}^infty I_k(a)I_{n-k}(b),
$$

in order to retrieve
$$
X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R) int_0^infty exp(-p r) I_{k}(omega r), dr,
$$

which is a doubly.infinite series of Laplace transforms,
$$
X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R), L[I_{k}(omega, cdot)](p),
$$

and, thanks to the property $I_k(cdot)=I_{-k}(cdot)$, this expression factorizes with respect to $L[I_0(omega, cdot)](p)=1/sqrt{p^2 - omega^2}$ as follows,
$$
X_n(R)=frac{1}{sqrt{p^2 - omega^2}} sum_{k=-infty}^infty I_{n-k}(omega R), left(frac{omega}{p+sqrt{p^2 - omega^2}} right)^{|k|}.
$$

At this point, I was hoping to use the generating function to get rid of the series, at least in the simplest case $n=0$, but having the absolute value $|k|$ cancels all the negative powers, and makes this approach inconclusive. Especially because someone aleady asked for any good value of "half the sum" of the generating function here.



Did I miss anything ? or did i make errors in my derivations ? Any help on this delicate question will be greatly appreciated ... Happy new year 2019 to everyone!










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    0














    Dear all: i'm trying to derive a closed form the following integral,
    $$
    X_n(R)=int_0^infty exp(-p, r)I_n(omega (r+R)), dr,
    $$

    where $I_n$ is the standard modified Bessel function of the first kind and $p>omega$, $0 leq omega<1$. I know that, when there is no shift ($R=0$), the values $X_n(R=0)$ are well-known because they are standard Laplace transforms $L[I_n(omega, cdot)](p)$, see for instance here.



    Yet, for $R not =0$, i've come up with 2 possibilities,



    -- using a "shift theorem" for the Laplace transform, which means that i change variable $r to r+R$ in order to get an exponential term $exp(pR)$ outside. However, this has an effect on the interval of integration,
    $$
    X_n(R)=exp(pR) int_R^infty exp(-pr)I_n(omega, r), dr,
    $$

    so that the value of the integral isn't well-known anymore. If $|R| ll 1$, one may use a polynomial approximation of $I_n$ in order to (approximately) compute the term
    $$
    Y_n(R)=int_0^R exp(-pr)I_n(omega, r), dr simeq int_0^R exp(-pr)frac{x^n}{2^n ,n!} , dr,
    $$

    which should be subtracted from the value of the Laplace transform as follows:
    $$
    X_n(R)=exp(pR)Big(L[I_n(omega, cdot)](p)-Y_n(R) Big).
    $$



    -- using the "summation formula" for Bessel functions,
    $$
    I_n(a+b)=sum_{k=-infty}^infty I_k(a)I_{n-k}(b),
    $$

    in order to retrieve
    $$
    X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R) int_0^infty exp(-p r) I_{k}(omega r), dr,
    $$

    which is a doubly.infinite series of Laplace transforms,
    $$
    X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R), L[I_{k}(omega, cdot)](p),
    $$

    and, thanks to the property $I_k(cdot)=I_{-k}(cdot)$, this expression factorizes with respect to $L[I_0(omega, cdot)](p)=1/sqrt{p^2 - omega^2}$ as follows,
    $$
    X_n(R)=frac{1}{sqrt{p^2 - omega^2}} sum_{k=-infty}^infty I_{n-k}(omega R), left(frac{omega}{p+sqrt{p^2 - omega^2}} right)^{|k|}.
    $$

    At this point, I was hoping to use the generating function to get rid of the series, at least in the simplest case $n=0$, but having the absolute value $|k|$ cancels all the negative powers, and makes this approach inconclusive. Especially because someone aleady asked for any good value of "half the sum" of the generating function here.



    Did I miss anything ? or did i make errors in my derivations ? Any help on this delicate question will be greatly appreciated ... Happy new year 2019 to everyone!










    share|cite|improve this question

























      0












      0








      0







      Dear all: i'm trying to derive a closed form the following integral,
      $$
      X_n(R)=int_0^infty exp(-p, r)I_n(omega (r+R)), dr,
      $$

      where $I_n$ is the standard modified Bessel function of the first kind and $p>omega$, $0 leq omega<1$. I know that, when there is no shift ($R=0$), the values $X_n(R=0)$ are well-known because they are standard Laplace transforms $L[I_n(omega, cdot)](p)$, see for instance here.



      Yet, for $R not =0$, i've come up with 2 possibilities,



      -- using a "shift theorem" for the Laplace transform, which means that i change variable $r to r+R$ in order to get an exponential term $exp(pR)$ outside. However, this has an effect on the interval of integration,
      $$
      X_n(R)=exp(pR) int_R^infty exp(-pr)I_n(omega, r), dr,
      $$

      so that the value of the integral isn't well-known anymore. If $|R| ll 1$, one may use a polynomial approximation of $I_n$ in order to (approximately) compute the term
      $$
      Y_n(R)=int_0^R exp(-pr)I_n(omega, r), dr simeq int_0^R exp(-pr)frac{x^n}{2^n ,n!} , dr,
      $$

      which should be subtracted from the value of the Laplace transform as follows:
      $$
      X_n(R)=exp(pR)Big(L[I_n(omega, cdot)](p)-Y_n(R) Big).
      $$



      -- using the "summation formula" for Bessel functions,
      $$
      I_n(a+b)=sum_{k=-infty}^infty I_k(a)I_{n-k}(b),
      $$

      in order to retrieve
      $$
      X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R) int_0^infty exp(-p r) I_{k}(omega r), dr,
      $$

      which is a doubly.infinite series of Laplace transforms,
      $$
      X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R), L[I_{k}(omega, cdot)](p),
      $$

      and, thanks to the property $I_k(cdot)=I_{-k}(cdot)$, this expression factorizes with respect to $L[I_0(omega, cdot)](p)=1/sqrt{p^2 - omega^2}$ as follows,
      $$
      X_n(R)=frac{1}{sqrt{p^2 - omega^2}} sum_{k=-infty}^infty I_{n-k}(omega R), left(frac{omega}{p+sqrt{p^2 - omega^2}} right)^{|k|}.
      $$

      At this point, I was hoping to use the generating function to get rid of the series, at least in the simplest case $n=0$, but having the absolute value $|k|$ cancels all the negative powers, and makes this approach inconclusive. Especially because someone aleady asked for any good value of "half the sum" of the generating function here.



      Did I miss anything ? or did i make errors in my derivations ? Any help on this delicate question will be greatly appreciated ... Happy new year 2019 to everyone!










      share|cite|improve this question













      Dear all: i'm trying to derive a closed form the following integral,
      $$
      X_n(R)=int_0^infty exp(-p, r)I_n(omega (r+R)), dr,
      $$

      where $I_n$ is the standard modified Bessel function of the first kind and $p>omega$, $0 leq omega<1$. I know that, when there is no shift ($R=0$), the values $X_n(R=0)$ are well-known because they are standard Laplace transforms $L[I_n(omega, cdot)](p)$, see for instance here.



      Yet, for $R not =0$, i've come up with 2 possibilities,



      -- using a "shift theorem" for the Laplace transform, which means that i change variable $r to r+R$ in order to get an exponential term $exp(pR)$ outside. However, this has an effect on the interval of integration,
      $$
      X_n(R)=exp(pR) int_R^infty exp(-pr)I_n(omega, r), dr,
      $$

      so that the value of the integral isn't well-known anymore. If $|R| ll 1$, one may use a polynomial approximation of $I_n$ in order to (approximately) compute the term
      $$
      Y_n(R)=int_0^R exp(-pr)I_n(omega, r), dr simeq int_0^R exp(-pr)frac{x^n}{2^n ,n!} , dr,
      $$

      which should be subtracted from the value of the Laplace transform as follows:
      $$
      X_n(R)=exp(pR)Big(L[I_n(omega, cdot)](p)-Y_n(R) Big).
      $$



      -- using the "summation formula" for Bessel functions,
      $$
      I_n(a+b)=sum_{k=-infty}^infty I_k(a)I_{n-k}(b),
      $$

      in order to retrieve
      $$
      X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R) int_0^infty exp(-p r) I_{k}(omega r), dr,
      $$

      which is a doubly.infinite series of Laplace transforms,
      $$
      X_n(R)=sum_{k=-infty}^infty I_{n-k}(omega R), L[I_{k}(omega, cdot)](p),
      $$

      and, thanks to the property $I_k(cdot)=I_{-k}(cdot)$, this expression factorizes with respect to $L[I_0(omega, cdot)](p)=1/sqrt{p^2 - omega^2}$ as follows,
      $$
      X_n(R)=frac{1}{sqrt{p^2 - omega^2}} sum_{k=-infty}^infty I_{n-k}(omega R), left(frac{omega}{p+sqrt{p^2 - omega^2}} right)^{|k|}.
      $$

      At this point, I was hoping to use the generating function to get rid of the series, at least in the simplest case $n=0$, but having the absolute value $|k|$ cancels all the negative powers, and makes this approach inconclusive. Especially because someone aleady asked for any good value of "half the sum" of the generating function here.



      Did I miss anything ? or did i make errors in my derivations ? Any help on this delicate question will be greatly appreciated ... Happy new year 2019 to everyone!







      laplace-transform generating-functions bessel-functions






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      asked Jan 4 at 10:37









      LaurentLaurent

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          Well, we know that the standard modified Bessel function of the first kind is:



          $$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
          $$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$



          Now, for the Laplace transform we know that:



          $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
          $$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
          $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$



          Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):



          $$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$



          So:



          $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
          $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$






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            Well, we know that the standard modified Bessel function of the first kind is:



            $$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
            $$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$



            Now, for the Laplace transform we know that:



            $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
            $$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
            $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$



            Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):



            $$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$



            So:



            $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
            $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$






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              0














              Well, we know that the standard modified Bessel function of the first kind is:



              $$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
              $$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$



              Now, for the Laplace transform we know that:



              $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
              $$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
              $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$



              Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):



              $$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$



              So:



              $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
              $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$






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                0












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                Well, we know that the standard modified Bessel function of the first kind is:



                $$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
                $$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$



                Now, for the Laplace transform we know that:



                $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
                $$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
                $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$



                Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):



                $$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$



                So:



                $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
                $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$






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                Well, we know that the standard modified Bessel function of the first kind is:



                $$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
                $$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$



                Now, for the Laplace transform we know that:



                $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
                $$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
                $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$



                Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):



                $$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$



                So:



                $$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
                $$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$







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                answered Jan 4 at 11:06









                JanJan

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