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Summary:

I struggle to find an appropriate inverse Laplace transform of the following
$$ 2^n a^n \frac{p^{n1}}{(p+a)^{2n}}, \quad a>0$$
Main Question or Discussion Point
I struggle to find an appropriate inverse Laplace transform of the following
$$F(p)= 2^n a^n \frac{p^{n1}}{(p+a)^{2n}}, \quad a>0.$$
WolframAlpha gives as an answer
$$f(t)= 2^n a^n t^n \frac{_1F_1 (2n;n+1;at)}{\Gamma(n+1)}, \quad (_1F_1  \text{confluent hypergeometric function})$$
which seem to work for my original problem, but only for small ##n## (<50). Whenever I go close or above 50 (I use Matlab to plot things) the outcome gets bizarre (##n=50##):
although the actual simulation result (the process which is described by ##F(p)##) looks as it meant to be, smoothly getting to the shape on the following figure (##n=49##).
Therefore I conclude there is something wrong with the ##f(t)## given out by Wolfram. Using the Laplace transform handbook (russian one, idk if it's reasonable to link) the best I could do was$$\hat{F}(p)= 2^n a^n \frac{p^{n}}{(p+a)^{2n}} \div \frac{2^n a^n}{\Gamma(2n)}\frac{d^{n1}}{dt^{n1}}(e^{at}t^{2n1})=\hat{f}(t), $$
since (as the handbook suggests)
$$ \left( \frac{p}{p+1} \right)^{n+1}\frac{1}{(p+1)^a}\div \frac{n!}{\Gamma(n+a+1)}e^{t}t^a L_n^a(t), \ L_n^a(t)=\frac{e^tt^{a}}{n!}\frac{d^n}{dt^n} (e^{t}t^{n+a}),$$
which I'm not sure is legitimate to use (because of ##n+1## power), yet it yields somewhat fine solution for my original problem (which is strange, because obviously ##F(p)\ne \hat{F}(p)##). I tried to plot it, and for small ##n## it is fine, but Matlab refuses to calculate the derivative (it spits out 0) for ##n>30##, so I cannot check whether it really suits. So my question is whether the Wolfram answer is wrong. I'm not really familiar with the confluent hypergeometric function, so I don't know how to interpret the outcome. What would you suggest to calculate the ##F(p)##? Is it even possible to do in elementary functions? I'd appreciate any help in the solution. I am aware of possibility that my reasoning is wrong, so in case of that let me know. Thank you for help!
$$F(p)= 2^n a^n \frac{p^{n1}}{(p+a)^{2n}}, \quad a>0.$$
WolframAlpha gives as an answer
$$f(t)= 2^n a^n t^n \frac{_1F_1 (2n;n+1;at)}{\Gamma(n+1)}, \quad (_1F_1  \text{confluent hypergeometric function})$$
which seem to work for my original problem, but only for small ##n## (<50). Whenever I go close or above 50 (I use Matlab to plot things) the outcome gets bizarre (##n=50##):
since (as the handbook suggests)
$$ \left( \frac{p}{p+1} \right)^{n+1}\frac{1}{(p+1)^a}\div \frac{n!}{\Gamma(n+a+1)}e^{t}t^a L_n^a(t), \ L_n^a(t)=\frac{e^tt^{a}}{n!}\frac{d^n}{dt^n} (e^{t}t^{n+a}),$$
which I'm not sure is legitimate to use (because of ##n+1## power), yet it yields somewhat fine solution for my original problem (which is strange, because obviously ##F(p)\ne \hat{F}(p)##). I tried to plot it, and for small ##n## it is fine, but Matlab refuses to calculate the derivative (it spits out 0) for ##n>30##, so I cannot check whether it really suits. So my question is whether the Wolfram answer is wrong. I'm not really familiar with the confluent hypergeometric function, so I don't know how to interpret the outcome. What would you suggest to calculate the ##F(p)##? Is it even possible to do in elementary functions? I'd appreciate any help in the solution. I am aware of possibility that my reasoning is wrong, so in case of that let me know. Thank you for help!