$\int_0^\infty 2xe^{-2x} \: dx=Γ(2)2(1/2)^2$ how to find this result of the integral?

Question

$$\int_0^\infty 2xe^{-2x} \: dx=Γ(2)2(1/2)^2$$

I don't understand. How can we write this? Please can you explain this clearly?

Answer 1

Start with the definition of the Gamma function $$\Gamma(n) = \int_0^{\infty} \: t^{n-1} e^{-t} dt$$ Substitute $t=2x$ in the definition $$= \int_0^{\infty} \: (2x)^{n-1} e^{-2x}\: 2dx$$ To match the power of $x$, set $n=2$. $$ \Gamma(2)= \int_0^{\infty} \: 2x e^{-2x} \: 2dx$$ Divide both sides by 2.

Answer 2

$$\int_0^{\infty} dx \: x^n \exp{(-\alpha x)} = \frac{\Gamma(n+1)}{\alpha^{n+1}} $$

You should be able to see your result immediately. The integral may be derived through integration by parts.

Answer 3

Make the substitution $u=2x$ and look up the definition of the Gamma function.