$$\int_{-\infty}^{\infty} g(t)dt = \int_{0}^{\infty} \frac{\lambda^{\alpha}}{\Gamma(\alpha)} t^{\alpha - 1} e^{-\lambda t} dt = \frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\infty} (\lambda t)^{\alpha - 1} e^{-\lambda t} dt$$

Set $u = \lambda t$:

$$= \frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\infty} u^{\alpha - 1} e^{-u} du = \frac{\lambda}{\Gamma(\alpha)} {\Gamma(\alpha)} = \lambda$$

I've tried this proof several times and always end up with an extra $\lambda$. Each step seems pretty straightforward so I can't find what I'm overlooking.


In your substitution for $u = λt$ you did not correctly replace the $dt$ term: you only substituted $du$ for it. In fact, $du = λ dt$, which "consumes" the last remaining $λ$ left of the integral.

  • $\begingroup$ Can't believe I missed that... Thanks! $\endgroup$
    – Y. Sargis
    Nov 22 '16 at 18:03

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