If we take for example, the problem of $$\int e^x \sin x \quad dx$$

We use the integration by parts technique:

$$\int uv' = uv - \int vu'$$


$\begin{array}{l l} u = \sin x & \frac{dv}{\color{red}{dx}} = e^x\\ \frac{du}{\color{red}{dx}} = \cos x & v = e^x\\ \end{array}$

However, I've seen people break up the derivative notation by bringing the denominator $dx$ over to the the RHS:

$\begin{array}{l l} u = \sin x & dv = e^x \; \color{red}{dx}\\ du = \cos x \; \color{red}{dx} & v = e^x\\ \end{array}$

I don't understand why people do this. With integration by substitution, this manipulation of the derivative notation is justified. But here it serves no purpose.


2 Answers 2


When you write $\int uv' = uv - \int vu'$, the notation is incomplete. Assuming the variable of integration is $x$, you should have written: $$ \int uv'\;dx = uv - \int vu' \;dx$$ In that case, we need to have a $dx$ in each integral symbol. If $u=f(x)$ and $v=g(x)$, then $du = f'(x)\;dx$ and $dv=g'(x)\;dx$, so the IBP formula is written: $$ \int u\;dv = uv - \int v\;du$$ $$ \int f(x)g'(x)\;dx = f(x)g(x) - \int g(x)f'(x)\;dx$$ Think of the "$dx$" as part of the notation.

  • $\begingroup$ Shouldn't your last line have $f\color{red}'(x)$ instead of $f(x)$ on the far right integral? $\endgroup$
    – ptrcao
    Dec 6, 2011 at 3:58
  • $\begingroup$ Right you are @ptrcao! I've just updated it. $\endgroup$
    – Shaun Ault
    Dec 6, 2011 at 21:52

The intuitive way of thinking about integration is all about summing differentials, and the notation that is taught to everyone nowadays translates this intuition into notation, even though it is not rigor.

The reason for this is that when you learn integration for the first time, one usually learns Riemann integration because it handles most of the function a beginner in mathematics needs to integrate, plus it allows one to justify the study of sums and series, derivatives and such in a manner that is user-friendly (up to some point). I believe that one who wishes to get to rigor needs to go through this phase at some point and then go to rigor in more advanced courses, it's his baby steps.

Now the reason why I'm saying this is because it is true that one should consider $u$ and $v'$ instead of $u$ and $dv$ when applying the integration by parts formula, if we stick to the theorems that say that those things work. But if one thinks of $dv$ as a differential, what you're doing by considering the integration by parts formula is that you are switching the roles of the differentials and the height function in a tricky manner (the trick is just the expansion of (uv)'), hence it makes more sense to say that 'this height function becomes this differential, and this differential becomes this height' in the intuitive sense, i.e. you're not thinking about what happens analytically (derivatives are integrated and functions are integrated), but more of what happens geometrically. In the end it's just about a choice of notation, and for pedagogic reasons we stick to differentials.

Actually, you can define a differential in this manner : given a differential $dx$ and a differentiable function $f'(x)$, it makes sense to define the new differential $$ dy(x,dx) = dy = f'(x)dx. $$ The differential $dy$ becomes a function of $x$ and $dx$, and one readily sees that this definition gives us $dy/dx = f'(x)$, so if one sticks to differentials there's a way to make some theorems with them, but in the end I don't have faith that it's worth it that much, it's more for the intuition part that I still think it's worth introducing them, mostly because for integration, proving stuff becomes a pain.

Hope that helps,

  • 5
    $\begingroup$ Emphasis on rigor for the last couple of centuries has caused non-rigorous math to atrophy somewhat. If I say that $f(x)$ is in meters per second and $dx$ is in seconds, therefore $f(x)\;dx$ is in meters, I don't think anyone would think of disputing that for a second, but I think there are many mathtematicians who don't instantly think of that and even more who don't mention it when they teach calculus. $\endgroup$ Dec 6, 2011 at 2:19
  • 4
    $\begingroup$ I am a fan of rigor, but I am also a fan of intuition. Intuition leads you to rigor and rigor is the foundation of intuition. One doesn't go without the other ; you need both. If you stay forever in intuition, you won't prove anything... but if you stay forever in rigor you will be blind. $\endgroup$ Dec 6, 2011 at 5:26

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