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Given this formula as known:

        $\int_0^\infty y^ke^{-y} dy=k!$

I want to use it to solve:

        $\int_0^\infty y^xe^{-3y} dy$

So I dutifully manipulate the term such that it matches the formula:

        $\int_0^\infty \frac{3^x}{3^x} y^x e^{-3y} dy = \frac{1}{3^x} \int_0^\infty (3y)^x e^{-(3y)} dy$

And here's where I went wrong, my answer was the following, and I'm missing a $\frac{1}{3}$ term:

        $\frac{x!}{3^x}$

I believe my error is because I don't have $y^ke^{-y}$ I have $(f(y))^k e^{-f(y)}$

In the question here: Integrating $e^{f(x)}$ I see where the correct answer $\frac{1}{3}\frac{x!}{3^x}$ comes from. Though I also note the rather non-obvious constraint that $f(y)$ be linear.

My question is: given a formula like this, and recognizing the existence of $f(y)=3y$, without necessarily knowing how the formula was derived (it's hard to understand every detail of every formula you might come across), how do I derive the extra $\frac{1}{3}$ term that I failed to identify in my first pass?

In the context that I came across this question, this seemed to be reasonably expected. How do I learn from my mistake here?

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    $\begingroup$ $\int_0^\infty \frac{3^x}{3^x} y^x e^{-3y} dy, u = 3y, du = 3\ dy \implies \int_0^\infty (\frac {1}{3^x}) u^x e^{-u} (\frac 13) du$ Hopefully this explains where your $\frac 13$ factor disappeared to. $\endgroup$ – Doug M Dec 11 '17 at 20:05
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    $\begingroup$ Ahh, so if I simply do the u-substitution the term pops out and I can safely apply the formula. So the net of it is that if I have f(y) applied to a formula where just y is presented, I should always take the extra step to apply u-substitution and the correct answer falls out naturally. This makes sense now and I think it's the key I was missing here. Thank you. @DougM Please do post that as an answer, that's what I needed to understand. $\endgroup$ – David Parks Dec 11 '17 at 20:11
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Hint:

$$ dy=\frac{1}{3} d(3y) $$

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