Prove that $$ \int_{0}^{\infty}x^{2019}\sin(\sqrt{3}x)e^{-3x}\mathrm{d}x=\dfrac{2019!\sqrt{3}}{2^{2021}\cdot 3^{1010}} $$
Hence generalize the integral for any other value than $2019$.I know it can be done by considering the integral $\displaystyle \int_{0}^{\infty} \sin(ax)e^{-bx}\mathrm{d}x=\dfrac{a}{a^2+b^2}$ and differentiating it again and again with respect to $b$, but it only works for smaller values, after some time calculation becomes very tedious.
Hint:
We integrate by parts $$\int_0^\infty x^ne^{wx}dx$$ where $w:=a+ib$ and $a<0$.
$$I_n:=\int_0^\infty x^ne^{wx}dx=\frac1w\left.x^ne^{wx}\right|_0^{\infty}-\frac nw\int_0^\infty x^{n-1}e^{wx}dx=-\frac nwI_{n-1}.$$
Then by induction,
$$I_n=(-1)^n\frac{n!}{w^n}I_0=(-1)^{n+1}\frac{n!}{w^{n+1}}I_0.$$
From this, taking the imaginary part,
$$\int_0^\infty x^n\sin(bx)e^{ax}dx=(-1)^{n+1}\frac{n!\sin\left((n+1)\arctan\dfrac ba\right)}{(a^2+b^2)^{(n+1)/2}}.$$
Here, $n=2019,a=-3,b=\sqrt3$.