Solving the Laplace transform $mathcal{L}_tleft{sinleft(at^nright)right}left(sright)$












2












$begingroup$


I was solving an integral and i stepped in a Laplace transormation of the form $mathcal{L}_tleft{sinleft(at^nright)right}left(sright)$ and I was curious on a generalized solution. After some work I wrote the following identity:



edit: sorry, I made a terrible mistake, the right identity is:
$$mathcal{L}_tleft{sinleft(at^nright)right}left(sright)=frac{a}{s^{n+1}}sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)}$$



where



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sinleft(aright)}{a} $ for $n=0$ (trivial)



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{s^2}{a^2+s^2} $ for $n=1$ (geometric series)



and



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sqrt{pi}}{2}e^{-frac{a^2}{4s}} $ for $n=frac{1}{2}$ $left(Gammaleft(x+frac{3}{2}right)=frac{sqrt{pi}left(2x+1right)!}{2^{2x+1}x!}right)$



Do you know how to get a general closed formula of this series?






sequences-and-series improper-integrals laplace-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
    $endgroup$
    – Brevan Ellefsen
    Jan 8 at 21:43










  • $begingroup$
    How did you arrive at your identity?
    $endgroup$
    – user150203
    Jan 9 at 0:02






  • 1




    $begingroup$
    @DavidG Taylor expansion of the sine and geometric series
    $endgroup$
    – Fabio
    Jan 9 at 0:08






  • 1




    $begingroup$
    It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
    $endgroup$
    – Maxim
    Jan 17 at 14:44










  • $begingroup$
    @Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
    $endgroup$
    – Fabio
    Jan 17 at 14:53
















2












$begingroup$


I was solving an integral and i stepped in a Laplace transormation of the form $mathcal{L}_tleft{sinleft(at^nright)right}left(sright)$ and I was curious on a generalized solution. After some work I wrote the following identity:



edit: sorry, I made a terrible mistake, the right identity is:
$$mathcal{L}_tleft{sinleft(at^nright)right}left(sright)=frac{a}{s^{n+1}}sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)}$$



where



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sinleft(aright)}{a} $ for $n=0$ (trivial)



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{s^2}{a^2+s^2} $ for $n=1$ (geometric series)



and



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sqrt{pi}}{2}e^{-frac{a^2}{4s}} $ for $n=frac{1}{2}$ $left(Gammaleft(x+frac{3}{2}right)=frac{sqrt{pi}left(2x+1right)!}{2^{2x+1}x!}right)$



Do you know how to get a general closed formula of this series?






sequences-and-series improper-integrals laplace-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
    $endgroup$
    – Brevan Ellefsen
    Jan 8 at 21:43










  • $begingroup$
    How did you arrive at your identity?
    $endgroup$
    – user150203
    Jan 9 at 0:02






  • 1




    $begingroup$
    @DavidG Taylor expansion of the sine and geometric series
    $endgroup$
    – Fabio
    Jan 9 at 0:08






  • 1




    $begingroup$
    It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
    $endgroup$
    – Maxim
    Jan 17 at 14:44










  • $begingroup$
    @Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
    $endgroup$
    – Fabio
    Jan 17 at 14:53














2












2








2





$begingroup$


I was solving an integral and i stepped in a Laplace transormation of the form $mathcal{L}_tleft{sinleft(at^nright)right}left(sright)$ and I was curious on a generalized solution. After some work I wrote the following identity:



edit: sorry, I made a terrible mistake, the right identity is:
$$mathcal{L}_tleft{sinleft(at^nright)right}left(sright)=frac{a}{s^{n+1}}sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)}$$



where



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sinleft(aright)}{a} $ for $n=0$ (trivial)



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{s^2}{a^2+s^2} $ for $n=1$ (geometric series)



and



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sqrt{pi}}{2}e^{-frac{a^2}{4s}} $ for $n=frac{1}{2}$ $left(Gammaleft(x+frac{3}{2}right)=frac{sqrt{pi}left(2x+1right)!}{2^{2x+1}x!}right)$



Do you know how to get a general closed formula of this series?






sequences-and-series improper-integrals laplace-transform







share|cite|improve this question











$endgroup$




I was solving an integral and i stepped in a Laplace transormation of the form $mathcal{L}_tleft{sinleft(at^nright)right}left(sright)$ and I was curious on a generalized solution. After some work I wrote the following identity:



edit: sorry, I made a terrible mistake, the right identity is:
$$mathcal{L}_tleft{sinleft(at^nright)right}left(sright)=frac{a}{s^{n+1}}sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)}$$



where



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sinleft(aright)}{a} $ for $n=0$ (trivial)



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{s^2}{a^2+s^2} $ for $n=1$ (geometric series)



and



$sum_{k in mathbb{N}_0}{left(frac{(-1)^kleft(a^2s^{-2n}right)^k}{left(2k+1right)!}Gammaleft(2kn+n+1right)right)} = frac{sqrt{pi}}{2}e^{-frac{a^2}{4s}} $ for $n=frac{1}{2}$ $left(Gammaleft(x+frac{3}{2}right)=frac{sqrt{pi}left(2x+1right)!}{2^{2x+1}x!}right)$



Do you know how to get a general closed formula of this series?






sequences-and-series improper-integrals laplace-transform




sequences-and-series improper-integrals laplace-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 23:14







Fabio

















asked Jan 8 at 18:18









FabioFabio

1079




1079








  • 1




    $begingroup$
    Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
    $endgroup$
    – Brevan Ellefsen
    Jan 8 at 21:43










  • $begingroup$
    How did you arrive at your identity?
    $endgroup$
    – user150203
    Jan 9 at 0:02






  • 1




    $begingroup$
    @DavidG Taylor expansion of the sine and geometric series
    $endgroup$
    – Fabio
    Jan 9 at 0:08






  • 1




    $begingroup$
    It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
    $endgroup$
    – Maxim
    Jan 17 at 14:44










  • $begingroup$
    @Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
    $endgroup$
    – Fabio
    Jan 17 at 14:53














  • 1




    $begingroup$
    Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
    $endgroup$
    – Brevan Ellefsen
    Jan 8 at 21:43










  • $begingroup$
    How did you arrive at your identity?
    $endgroup$
    – user150203
    Jan 9 at 0:02






  • 1




    $begingroup$
    @DavidG Taylor expansion of the sine and geometric series
    $endgroup$
    – Fabio
    Jan 9 at 0:08






  • 1




    $begingroup$
    It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
    $endgroup$
    – Maxim
    Jan 17 at 14:44










  • $begingroup$
    @Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
    $endgroup$
    – Fabio
    Jan 17 at 14:53








1




1




$begingroup$
Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
$endgroup$
– Brevan Ellefsen
Jan 8 at 21:43




$begingroup$
Let's clean up a bit to make the answer a little more obvious. Let $alpha = -a^2$ and $z = s^{2n}.$ You wish to know if we can analytically continue the sum $$sum_{k ,in ,mathbb{N}} left(frac{alpha}{z}right)^k$$ Which should have a meromorphic extension away from any poles (which will of course depend on $n,$ since $z$ does) by the standard formula for a geometric series.
$endgroup$
– Brevan Ellefsen
Jan 8 at 21:43












$begingroup$
How did you arrive at your identity?
$endgroup$
– user150203
Jan 9 at 0:02




$begingroup$
How did you arrive at your identity?
$endgroup$
– user150203
Jan 9 at 0:02




1




1




$begingroup$
@DavidG Taylor expansion of the sine and geometric series
$endgroup$
– Fabio
Jan 9 at 0:08




$begingroup$
@DavidG Taylor expansion of the sine and geometric series
$endgroup$
– Fabio
Jan 9 at 0:08




1




1




$begingroup$
It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
$endgroup$
– Maxim
Jan 17 at 14:44




$begingroup$
It's the Fox-Wright function. Note that the sum diverges for $n > 1$.
$endgroup$
– Maxim
Jan 17 at 14:44












$begingroup$
@Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
$endgroup$
– Fabio
Jan 17 at 14:53




$begingroup$
@Maxim At the beginning I thought that the series would diverge for $n>1$ too, but wolfram can give exact results in form of Fresnel integrals for $n=2$ or $n=4$ as far as I checked
$endgroup$
– Fabio
Jan 17 at 14:53










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