i am trying to derive a generating function from a probability density function (PDF) on $x \geq 0$.
i think i'm supposed to apply the following formula to do this, where $f(x)$ is the PDF.
$\int_{0}^\infty e^{tx} \cdot f(x)dx$
the formula for my specific PDF is below.
$f(x) = \frac{e^{-x/3}}{3}$
so applying the formula from the beginning:
$\int_{0}^\infty e^{tx} \cdot \frac{e^{-x/3}}{3}dx$
i know that the answer is $\frac{1}{1-3t}$, but i am having a lot of trouble with the integration process. i have tried u substitution and integration by parts, but i just can't seem to crack it.
any advice for how to proceed? thank you in advance.
Note that $e^{tx}\cdot e^{-x/3}=e^{x(t-1/3)}$.
Hence, $\int_{0}^\infty e^{tx} \cdot \frac{e^{-x/3}}{3}dx=\frac 13\int_{0}^\infty e^{x(t-1/3)}dx$. Now use $u$ substitution with $u=x(t-1/3)$. Treat $t$ as just a number.
Of course, you need $t<1/3$ for the integral to converge.