Normalizing an exponential distribution

Question

I need some help to solve this problem:

Evaluate A such that the exponential distribution with parameter $\alpha, P(X = x) = Ae^{−\alpha x}$, is normalized. Here, $\alpha > 0$ and $\Omega = \mathbb{R}_{+}$.

I've been trying to evaluate the following Integral

$$ \int^{\infty}_{0}Ae^{-\alpha x}dx=1 \,\, , $$

I always get teh result that $A$ must be equal to 0... Am I making something wrong?

Edit: What I did:

$$ \int^{\infty}_{0}Ae^{-\alpha x}dx=1 \Longrightarrow -\frac{A}{\alpha}\lim_{b\to \infty} \int^{b}_{0}e^{-\alpha x}dx=1 \,\, , $$

Since $\lim_{b\to \infty}e^{-\alpha \cdot \infty}=0$, I get the not true equality

$$ 0=1 \,\, . $$

Answer 1

Note that $$\int_{0}^{\infty} e^{-\alpha x} \, \mathrm{d}x = -\frac{1}{\alpha}\bigg[e^{-\alpha x}\bigg]_0^{\infty} = -\frac{1}{\alpha} (0 - 1) = \frac{1}{\alpha}$$since $\lim_{x\to \infty} e^{-\alpha x} = 0$ for $\alpha > 0$ and $e^0 = 1$. This gives $A = \alpha$.

Answer 2

HINT $$ \int_0^\infty Ae^{-ax}dx = A \int_0^\infty e^{-ax}dx = \left. \frac{Ae^{-ax}}{-a} \right|_0^\infty = \frac{A}{a} $$