Evaluating $\Gamma$-function at $x=2$

Question

We know from lecture that the $\Gamma$-function has the property $\Gamma(2)=\Gamma(1)=1$. If I evaluate the corresponding integral: $$\Gamma(x)=\int^{\infty}_0 t^{x-1}e^{-t} dt$$ at $x=1$ then I get: $\Gamma(1)=1$. So far so good.

However, if I evaluate the integral at $x=2$ I don't get $\Gamma(2)=1$.

This is my calculation: $$\Gamma(2)=\int^{\infty}_0 t^{2-1}e^{-t} dt=\int^{\infty}_0 te^{-t} dt=-te^{-t}\vert^{\infty}_0-\int^{\infty}_0e^{-t}dt =-te^{-t}\vert^{\infty}_0-1 =-1 $$

Maybe someone can tell me where I am wrong.

Answer 1

You dropped a minus sign. After the integration by parts, the integral in the second term gets a minus sign both from the anti-derivative of $\mathrm e^{-t}$ and from the minus sign in the rule for integration by parts; so it should be positive.

Answer 2

@joriki is right about their being a sign error. To evaluate $\int_0^\infty te^{-t}dt$ as $\int_0^\infty uv^\prime dt$ take $u=t,\,v=-e^{-t}$ so $-\int_0^\infty u^\prime vdt=\int_0^\infty e^{-t}dt$.