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I'm currently doing integration by parts, and I'm finding that the notation is what makes it tough for me. So I looked it up and found that:

$$\int u(x)v'(x)dx= u(x)v(x) - \int u'(x)v(x)dx$$

But the wikipedia said that this above equality is the same as:

$$ \int u dv = uv - \int v du $$

I think I have located my problem: I don't see how these 2 are equal at all. How could for example know that $$\int u(x)v'(x) dx = \int udv$$

I can't really see the equivalence here.

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The point is that when Leibniz conceived the derivative, he thought of it as the quotient of two infinitely small numbers $dy$ and $dx$ if $y = f(x)$, for example. In that case, the derivative was really written as $dy/dx$, and since $f'(x)=dy/dx$, we have $dy=f'(x)dx$ as the infinitely small quantity that $y$ varies at a rate $f'(x)$ when $x$ varies the infinitely small quantity $dx$.

However, this stuff isn't rigorous. Indeed, in standard analysis, it is impossible to conceive a number like an infinitesimal, and the use of this even as mere notation may lead to confusion. That's why in the modern language, we simply use the prime notation to denote the derivative. The next best thing to replace the infinitesimals $dy$ and $dx$ is the notion of a differential form; there's so much about them to be said that I won't explain here.

So, in truth, if you use this stuff that's not rigorous you have $dv=v'(x)dx$ and $du=u'(x)dx$ so that we can write:

$$\int u(x)v'(x)dx = u(x)v(x)-\int v(x)u'(x)dx$$

Simply as:

$$\int udv = uv - \int v du$$

Now, understand that this last thing is just a mnemonic rule so that people remember what to do when find an integral like that. The rigorous version is the first formula that's derived directly from the product rule for derivatives.

Indeed, many of the notations regarding one-dimensional integrals that refer to "let $u$ be that, then $du$ is that other thing" and so forth are just rules for you to remember easier what to do. As I've said, those $dx$, $du$ and everything else can have a rigorous meaning as differential forms when you study differential geometry.

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  • $\begingroup$ The abbreviated equation actually makes sense in terms of the Riemann-Stieltjes integral. $\endgroup$ – Joseph Calland Jun 2 '13 at 18:26
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It is really bad shorthand notation that is pervasive throughout calculus that leads to a lot of handwavy mathematics. Really the shorthand reads: $dv = v'(x) dx$.

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  • $\begingroup$ So you would recommend my to use only the equality at the top to avoid confusion? $\endgroup$ – tomtit Jun 1 '13 at 21:15
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    $\begingroup$ As long as you remember what the shorthand means, use it. If you don't like it, you don't have to. It's just there to make expressions a little easier on the eye. $\endgroup$ – Cameron Williams Jun 1 '13 at 21:17
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You know $\frac {dv}{dx}=v'(x)$. Now if we treat $\frac {dv}{dx}$ as a fraction (even though it isn't) and multiply both sides by $dx$ (known as a differential), we get $dv=v'(x)dx$. Can you figure out the rest?

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The "answer" lies in the proof of Integration by Parts. It also helps to recall that $v^\prime(x)=\frac{d v}{d x}$.

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