Confused about the solution to this integral 
I am confused as to how the integral:
$$\int_0^\infty \theta \cdot x^{-3/2}dx$$
was evaluated as: $$2 \cdot \theta^{1/2}$$ 
with $x^{-1/2}$ in the equation and $x = \infty$, shouldn't that set everything to $0$ ?

Since $\theta\ $ is constant he equation $\int_o^\infty\theta\cdot x^{-3/2} \, dx $ can be written as:
$$\theta\cdot\int_o^\infty\ x^{-3/2} \, dx$$
Now $$\int x^{n} \,dx = \dfrac{x^{n+1}}{n+1}+C$$
Therefore,
$$\int x^{-3/2} \, dx=\dfrac{x^{\dfrac{-3}2+1}}{\dfrac{-3}2+1}+C= \dfrac{x^{-1/2}}{\dfrac{-1}2}$$
$$\int_o^\infty\theta\cdot x^{-3/2} \, dx=-2\theta\cdot[x^{-1/2}]_o^\infty$$
Now from here, I assume that the lower limit is $\theta$ instead of $0$
In that case
$$\int_o^\infty\theta\cdot x^{-3/2} \, dx=-2\theta\cdot[x^{-1/2}]_\theta^\infty$$
$$=-2\theta\cdot[0-\theta^{-1/2}] (\, since \, \dfrac 1{x}_{x->\infty}=0)$$
$$=2\cdot\theta^{1/2}$$
So, the lower limit should be $\theta$ instead of $0$
