By first making a substitution, evaluate the integral in terms of the gamma function
$$\int\limits_{1}^{∞}t^2e^{-t}.dt$$
the gamma function is defined
$$\Gamma(z)=\int\limits_{0}^{∞}e^{-x}x^{z-1}.dx$$ By first making a substitution, evaluate the integral in terms of the gamma function
$$\int\limits_{1}^{∞}t^2e^{-t}.dt$$
the gamma function is defined
$$\Gamma(z)=\int\limits_{0}^{∞}e^{-x}x^{z-1}.dx$$ @Mathematics

Question

zarkon · Answer

$$u=t-1$$

$$t=u+1$$

$$du=dt$$

$$\int\limits_{1}^{∞}t^2e^{-t}\,dt=\int\limits_{0}^{∞}(u+1)^2e^{-u-1}\,du$$

$$=e^{-1}\int\limits_{0}^{∞}(u^2+2u+1)e^{-u}\,du=e^{-1}(\Gamma(3)+2\Gamma(2)+\Gamma(1))$$
$$=e^{-1}(2+1+1)=5e^{-1}$$

zarkon · Answer

$$=e^{-1}(2+2+1)=5e^{-1}$$