I have tried mathematica to compute this integral, but I have found no answer.
$$ \int_{0}^{\infty} \exp(-\lambda x) x^{n} dx $$ for $\lambda > 0$.
I have a feeling that it is
$$ \frac{n!}{\lambda^{n+1}} $$ based on my assumption that the indefinite integral results in upper incomplete Gamma function and at the limits $0$ and $\infty$, the upper incomplete Gamma function results in complete Gamma function, and $0$, respectively. Please correct me if I am wrong.
If you know about gamma function, then the substitution $t = \lambda x$ gives you $$ \int\limits_0^\infty \exp(-\lambda x)x^n dx = \frac{1}{\lambda^{n+1}}\int\limits_0^\infty \exp(-t)t^n dt = \frac{\Gamma(n+1)}{\lambda^{n+1}} = \frac{n!}{\lambda^{n+1}}. $$ Thus, the same result can be easily obtained integrating by parts