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

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

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

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Is there a reason you write the constant as $2(1/2)^2$ instead of $1/2$? – Christopher A. Wong Jan 7 '13 at 20:21
Actually, I think writing it like that is helpful given the answer. – Ron Gordon Jan 7 '13 at 20:21

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.

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please also note that $\Gamma(n) = (n-1)!$ for positive integers – karakfa Jan 8 '13 at 15:27

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

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

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Yes! Thank you:) – B11b Jan 7 '13 at 20:21

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

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