1
$\begingroup$

It is common that we replace $\int u(x)v′(x)\mathrm{d}x$ by $\int u \mathrm{d} v$ where both $u$ and $v$ are continuous functions of $x$. My question is, must we ensure that $u$ can be written as a function of $v$ before applying this? The above substitution method is involved in the proof of integration by parts but I cannot find textbooks that addressed this point.

I find it easier to understand for definite integrals because I can think it as summation. But for indefinite integrals it is just anti-derivative and I am not sure how this kind of replacement is valid.

In terms of differential, of course $\mathrm{d}v$ and $v′(x) \mathrm{d}x$ are the same. But $∫$ and $\mathrm{d}x$ together forms one single mathematical sign meaning anti-derivative for indefinite integral, $∫f(x)\mathrm{d}x$ means finding the primitive function $F(x)$ which is a function of $x$. So to me $∫u\mathrm{d}v$ implies $u$ is function of $v$.

I just wonder why in applications substitution is done in this way without considering the differentiability of $u$ with respect to $v$.

$\endgroup$
  • $\begingroup$ Actually, $\int\mathrm{d}x$ isn't one single mathematical sign. You'll notice this in multivariable and vector calculus later. $\endgroup$ – UserX Nov 27 '14 at 6:12
0
$\begingroup$

Here's one way to think about it. Let's say $d$ and $\int$ are inverse operators of each other. The $d$ operation takes the differential of a function, so by definition $\frac{d(f(x))}{dx}=f'(x)$ and for example $d(x^2) = 2x dx$

Now if we say that the $\int$ operator is the inverse of that, then it stands that $\int du = u$ by definition.

Now here is where we can justify the substitution rule. Lets say you have an integral of this form: $\int f(x)g'(x)dx$. Since $g'(x)=\frac{d(g(x))}{dx}$, we can rewrite the integral as: $\int f(x)\frac{d(g(x))}{dx}dx$, and the $dx$ cancels, so you are left with $\int f(x)d(g(x))$

If we replace $f(x)$ with $u$ and $g(x)$ with $v$, then the integral is $\int u(x)v'(x)dx$ and we have just shown that that must be equal to $\int udv$

This is just one way of thinking about it that I really enjoy because it justifies substitution nicely.

To answer your original question, yes it does mean u is a function of v, simply because it can always be written like that. You can always solve for x and then write one function in terms of the other and its inverse.

Ex: let $u=f(x)$ and $v=g(x)$, then $g^{-1}(v)=x$ by the definition of an inverse function, and so $u=f(x)=f(g^{-1}(v))$ which is a function of v.

Hope you could understand this, let me know if any of it confuses you.

$\endgroup$
  • $\begingroup$ Thank you. Your answer is useful. What still confuses me is that why the dx cancels in ∫f(x)g'(x)dx? To me ∫dx is a single mathematical sign and the derivative is another, why they can be simply cancelled like those in fractions? Another question is, in complicated functions and implicit functions, can we be sure that u is a function of v by finding inverse function like that you just mentioned? $\endgroup$ – Kelvin S Nov 27 '14 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.