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Please can someone help me? I have tried to find the Laplace transform of the form: $$\int_{0}^{+\infty} (v+1)^{\nu}(2v+1)^{k}v^{\alpha} \exp(-pv), \mbox{ where }\alpha,\nu, k \mbox{ are integers }$$ I have searched on the two Books "Erdelyi. Tables of integral transforms, Vol I, 1954" and "Oberhettinger, L. Badii, Tables of Laplace transforms, 1973. But i didn't found this type of integral transform is there any reference or method in order to compute this formula above. Thanks and best regards.

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    $\begingroup$ the Laplace transform equals $\sum_{s=0}^\infty c_s (s+\alpha)!/p^{s+\alpha+1}$, where $c_s$ is the coefficient of $v^s$ in the multinomial expansion of $(v+1)^\nu(2v+1)^k$; this coefficient is some hypergeometric function, I don't think there is a closed form expression in terms of elementary functions; of course, for specific values of $\alpha,\nu,k$ the Laplace transform can be found with a simple computation as a polynomial in $p$ with integer coefficients. $\endgroup$ Mar 17, 2021 at 13:43
  • $\begingroup$ Thank you but i didn't understood your result. The sum that you gave is the Laplace transform $\int_{0}^{+\infty} (v+1)^{nu}(2v+1)^{k}v^{\alpha}exp(−pv)$? How did you found this sum and what did you mean by " the coefficient $c_s$ is some hypergeometric function"? Can we deduce by computation this hypergeometric function? $\endgroup$ Mar 17, 2021 at 22:37
  • $\begingroup$ I have worked it out below. $\endgroup$ Mar 18, 2021 at 9:30

2 Answers 2

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For $n,k\in\mathbb{N}$ and $\alpha\geq 0$ we can use the binomial expansion, $$(v+1)^{\nu}(2v+1)^{k}v^{\alpha}=\sum_{s=0}^{\nu+k}\sum_{q=0}^{s}{{\nu}\choose{q}} {{k}\choose{s-q}}2^{s-q}v^{s+\alpha} $$ $$\Rightarrow \int_{0}^{\infty} (v+1)^{\nu}(2v+1)^{k}v^{\alpha} e^{-pv}\,dv=\sum_{s=0}^{\nu+k}\sum_{q=0}^{s}{{\nu}\choose{q}} {{k}\choose{s-q}}2^{s-q}\frac{(s+\alpha)!}{p^{s+\alpha+1}}$$ $$\qquad=\sum_{s=0}^{\nu+k}2^s \binom{k}{s} \, _2F_1\left(-\nu,-s;k-s+1;\tfrac{1}{2}\right)\frac{(s+\alpha)!}{p^{s+\alpha+1}}.$$

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  • $\begingroup$ But if $\nu$ and $k$ are negative numbers then we can not apply the Newton formula? What about this case? $\endgroup$ Mar 18, 2021 at 12:11
  • $\begingroup$ The command of Mathematica 12.2 Table[LaplaceTransform[(v + 1)^\[Nu]*(2 v + 1)^k*v^\[Alpha], v, p, Assumptions -> {\[Nu], \[Alpha]} \[Element] Integers && \[Alpha] >= 0, GenerateConditions -> False], {k, 2, 5}] performs answers in terms of $\text{Hypergeometric1F1}$ functions and $\Gamma$-functions. $\endgroup$
    – user64494
    Apr 17, 2021 at 16:04
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Another way to look at this integral is by considering $$F(q,r,p) = \int_{0}^{\infty}\exp\left[-q(v+1)-r(2v+1)-pv\right]\,dv = \frac{\exp(-q-r)}{p+q+2r}$$ for $p+q+2r>0$. If $\nu\geq 0$, $k\geq 0$, and $\alpha\geq 0$, then your integral can be represented as $$(-1)^{\nu+k+\alpha}\left.\frac{\partial^{\nu+k+\alpha}F}{\partial q^{\nu}\,\partial r^{k}\,\partial p^{\alpha}}\right|_{q\,=\,r\,=\,0}.$$ For negative values of $\nu$ or $k$, integration should be used instead of differentiation.

Of course, it depends on what you're looking for, so this method may not give anything better than the previous answer.

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  • $\begingroup$ For negatve integers values of $\alpha$ the Laplace transform under consideration does not exist. $\endgroup$
    – user64494
    Apr 18, 2021 at 3:54

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