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I want to solve the following integral $$\int_0^{\infty}y^3\theta e^{-\theta y} dy$$ so I chose two approaches, a direct one and one with substitution. The direct one is just a triple integration per parts which leads to $\frac{6}{\theta^3}$. The one with the substitution is solved by using the substitution $y\theta = t$. Now we have $y = \frac{t}{\theta}$ and $dy = \frac{dt}{\theta}$. Substituting and using per parts, we have $$\frac{1}{\theta^3}\int_0^{\infty}t^3e^{-t}dt$$ and we get again $\frac{6}{\theta^3}$.

However this is a result in $t$, not $y$ (the original variable). So I thought I had to find it in terms of $y$, i.e. $y = \frac{t}{\theta} = \frac{6}{\theta^4}$ which is different from the previous result!

So my question is

When I solve an integral using substitution as above, do I have to bring back the result in terms of the original variable? Cause I recall loads of examples (e.g. the sine substitution) where we had to bring the result back. But here obviously I get the same result if I don't bring it back to the original variable. Can you help me clarify this?

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    Nope, the result is independent of any variables you might have used before in order to obtain it. And the constant in front of the second integral is wrong, should be $\frac{1}{\theta^4}$ – 2017-01-08
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    @john_jerome the other one coming from $\frac{dt}{\theta}$ cancels out with the $\theta$ already in the integral – 2017-01-08
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    Sorry, my bad, you're right – 2017-01-08

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When the given integral is a definite integral, then there is no need to bring back to the initial variable, since we know that with an appropriate change of variable we have the equality $$ \int_a^bf(x)\:dx=\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(t))\:\phi'(t)\:dt. $$

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    thanks ! what about a indefinite integral? Do we have then? – 2017-01-08
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    where do we get the $\phi'(t)$ from? – 2017-01-08
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    @Euler_Salter You are welcome. Concerning an indefinite integral, one usually has to bring back the result to the original variable. – 2017-01-08
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    We have $x=\phi(t)$, $dx=\phi'(t)dt$. – 2017-01-08