Integration by substitution, rigorously

Dear Uncle Colin,

Can you explain why integration by substitution works? I get that you’re not allowed to ‘cancel’ the $dx$s, but can’t see how it works otherwise.

- Reasonable Interpretation Got Our Understanding Ridiculed

Hi, RIGOUR, and thanks for your message.

First up, confession time: me and rigour go together like peanut butter and lawnmowers. We are not a natural combination, and any attempt to apply one to the other is likely to end in a mess - or at least anguished screams from @realityminus3. But, since you asked, I’ll do my best.

It helps if we go back to what integration means: $\int y \dx$ gives the family of functions, $F(x)$, such that $\diff {F(x)}{x} = y$.

Also, $x$ is a dummy variable - it wouldn’t matter if you changed it for $t$ or $\theta$ or anything else - it’s just there as a placeholder. In particular, $\int y \d t$ gives a family of functions $F(t)$ which are (up to replacing the variable) the same ones as you’d get from integrating with respect to $x$.

Suppose we have $F(x)$, where $x$ is a function of $t$. Then, if we differentiate $F(x)$ with respect to $t$, we get $\diff {F(x)}{x} \cdot \diff xt$. Integrating this with respect to $t$ - a dummy variable, remember - gets us back to $F(x)$.

So, we have $F(x) = \int y \diff xt \d t$ - but we originally had $F(x) = \int y \dx$, so $\int y \diff xt \d t = \int y \dx$ as required. $\blacksquare$.

Hope that helps!

- Uncle Colin

* Edited 2018-10-03 to fix a typo. Thanks, Adam!