Show the gamma density function integrates to 1

[Catchfire]

Homework Statement

Show the gamma density function integrates to 1.

Homework Equations

Assume α > 0, λ > 0, t > 0
g(t) = $\frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt}$
$\Gamma (α)= \int_0^∞ t^{α-1} e^{-t} dt$

The Attempt at a Solution

Show $\int_0^∞ \frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt} dt = 1$

$\int_0^∞ \frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt} dt$
= $\frac{λ^α}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-λt}dt$
= $\frac{λ^α}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-t}e^λdt$
= $\frac{λ^αe^λ}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-t}dt$
= $\frac{λ^αe^λ \Gamma (α)}{\Gamma (α)}$
= $λ^αe^λ$

...where did I lose the plot?

 

[Dick]

Homework Statement

Show the gamma density function integrates to 1.

Homework Equations

Assume α > 0, λ > 0, t > 0
g(t) = $\frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt}$
$\Gamma (α)= \int_0^∞ t^{α-1} e^{-t} dt$

The Attempt at a Solution

Show $\int_0^∞ \frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt} dt = 1$

$\int_0^∞ \frac{λ^α}{\Gamma (α)} t^{α-1}e^{-λt} dt$
= $\frac{λ^α}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-λt}dt$
= $\frac{λ^α}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-t}e^λdt$
= $\frac{λ^αe^λ}{\Gamma (α)} \int_0^∞ t^{α-1}e^{-t}dt$
= $\frac{λ^αe^λ \Gamma (α)}{\Gamma (α)}$
= $λ^αe^λ$

...where did I lose the plot?

You lost it when you said $e^{-λt}=e^{-t}e^λ$. That's not true. I would try the variable substitution $u=λt$.

 

[Catchfire]

Looks like I need to refresh myself on the laws of exponents.

That substitution did the trick, thanks.