I was taking integral of $e^{x}x^{n}$ (nEZ+) and noticed it follows recurrence of
$a_{n}=\int e^{x}x^{n}dx=x^{n}e^{x}-na_{n-1}$ where $a_{0}=e^{x}$
I was trying to find a closed form, $f(n)$, with no avail. Is there any techniques for finding closed form of recurrence and if it's possible to find a closed form for any recurrence relationship. Thanks
The usual approach is to do a couple, then find a pattern, and lastly prove that the pattern you educatedly guessed is the right one, usually by induction.
In this case, you noted that $$a_n=x^ne^x-na_{n-1}.$$ We immediately note that the "times $n$" will induce a factorial term, so we'll be looking out for those.
Pluggin in $n-1$ gives $$a_{n-1}=x^{n-1}e^x-(n-1)a_{n-2},$$ Which we now plug back into the first equation, obtaining $$a_n=x^ne^x-na_{n-1}=x^ne^x-n(x^{n-1}e^x-(n-1)a_{n-2})=x^ne^x-nx^{n-1}e^x+n(n-1)a_{n-2}.$$ It is quite clear that the next term will include a summand of the form $x^{n-2}e^x$, and that its coefficient is $n\times(n-1)=(-1)^2\frac{n!}{n-2!}$...
From this, it is natural to claim that $$a_n=\sum_{k=0}^{n-1}(-1)^k\frac{n!}{n-k!}x^{n-k}e^{-x}+(-1)^nn!a_0$$