5
$\begingroup$

If we are given a function of $x$, $a(x)$, how hard is it to find an $f(x)$ and $g(x)$ such that $$a(x)=f(x)g'(x)+f'(x)g(x)$$ For comparison, I'd like to know when this is easier than symbolically or numerically integrating $a(x)$.

I'd like to know, if possible, what general conditions allow us to efficiently find $f(x)$ and $g(x)$. I'm hoping this isn't too general a question. Additionally, I'd like to know the methods that allow us to do so.

  • 0
    You want to find $f$ and $g$ such that $a=(fg)'$? Could you give an example?2012-07-11
  • 0
    @PeterTamaroff: Let $a(x)=\cos^3(x)-\sin^2(x)\cos(x)$. Then we can pick an $f(x) = \sin(x)$ and $g(x)=\cos^2(x)$. It's easy to come up with an example if you start with $f(x)$ and $g(x)$. But I find it tough to go the other way.2012-07-12
  • 1
    It can't be any easier than antidifferentiating $a$, since, if you can do it, you can antidifferentiate $a$ just by writing down $fg$.2012-07-12
  • 2
    It's as hard as integrating. If you can find $f,g$, then $fg$ is an indefinite integral. If you can find an indefinite integral $f$, then you can use $g = 1$.2012-07-12
  • 0
    A enough condition on $a$ is continuity. In that case you can write $$a(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left(1\cdot \int_{x_0}^x a(t)\mathrm{d}t\right).$$ Take $f\equiv 1$ and $g=\int_{x_0}^x a(t)\mathrm{d}t$. If you do not allow constants functions, it's an involved problem, don't is it?2012-07-12
  • 0
    Four upvotes and I still cannot understand the question. :-(2012-07-12
  • 1
    I like to think of this as the same as applying integration by parts, since it comes from $d(uv) = d(u)v + u(dv)$.2012-08-13

2 Answers 2

1

Suppose $a(x)=x^2\cos x + 2x\sin x$.

Recognizing that as $f'(x)g(x)+ f(x)g'(x)$, with $f(x)= \sin x$ and $g(x)= x^2$, seems like the quickest way of finding the antiderivative of the whole expression.

0

Isn't this just integration by parts. Where

$$ \int a(x)\,{\rm d}x = f(x)g(x)= \int \frac{{\rm d}f(x) g(x)}{{\rm d}x}\,{\rm d}x = \int f(x)g'(x) \,{\rm d}x +\int f'(x)g(x) \,{\rm d}x $$ $$ = \int f(x) \,{\rm d}g +\int g(x) \,{\rm d}f $$

or

$$ \int u \;{\rm d}v = u\, v - \int v \;{\rm d}u $$

  • 0
    How do you find $f(x)$ and $g(x)$, without knowing the solution to the integral?2012-08-13
  • 0
    The way integration by parts works is to transform $\int a {\rm d}x$ into $\int u {\rm d}v$ and then computing the r.h.s. of the equality above.2012-08-13