Show the gamma density function integrates to 1

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SUMMARY

The gamma density function, defined as g(t) = (λ^α / Γ(α)) t^(α-1)e^(-λt), integrates to 1 for α > 0 and λ > 0. The integral ∫_0^∞ g(t) dt can be simplified using the substitution u = λt, leading to the conclusion that the integral equals 1. The error in the original attempt stemmed from incorrectly manipulating the exponential term e^(-λt) as e^(-t)e^λ, which is not valid. Correct application of variable substitution resolves the integration issue.

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Catchfire
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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?
 
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Catchfire said:

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##.
 
Last edited:
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Looks like I need to refresh myself on the laws of exponents.

That substitution did the trick, thanks.
 

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