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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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  • $\begingroup$ 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 $\frac{1}{\theta^4}$ $\endgroup$ – john_jerome Jan 8 '17 at 14:04
  • $\begingroup$ @john_jerome the other one coming from $\frac{dt}{\theta}$ cancels out with the $\theta$ already in the integral $\endgroup$ – Euler_Salter Jan 8 '17 at 14:11
  • $\begingroup$ Sorry, my bad, you're right $\endgroup$ – john_jerome Jan 8 '17 at 18:59
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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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  • $\begingroup$ thanks ! what about a indefinite integral? Do we have then? $\endgroup$ – Euler_Salter Jan 8 '17 at 14:12
  • $\begingroup$ where do we get the $\phi'(t)$ from? $\endgroup$ – Euler_Salter Jan 8 '17 at 14:16
  • $\begingroup$ @Euler_Salter You are welcome. Concerning an indefinite integral, one usually has to bring back the result to the original variable. $\endgroup$ – Olivier Oloa Jan 8 '17 at 14:17
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    $\begingroup$ We have $x=\phi(t)$, $dx=\phi'(t)dt$. $\endgroup$ – Olivier Oloa Jan 8 '17 at 14:17

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