# Name for integration by parts shortcut

1. Oct 13, 2009

### PhantomOort

Hi all. I've recently learned a shortcut for integration by parts, but don't know what it's called or where it comes from.

The trick is to find $$\lambda$$ such that $$f'' = \lambda f$$ and $$\mu$$ such that $$g'' = \mu g$$, providing both are constants and $$\lambda$$$$\neq$$$$\mu$$. Then $$\int$$f(x)g(x)dx = $$\frac{f'g-fg'}{\lambda-\mu}$$.

Can anyone tell me what this is called? Thanks.

2. Oct 13, 2009

### HallsofIvy

Staff Emeritus
Is $\lambda$ a number here? Then I can't imagine that "shortcut" being much good. It is only possible to find $\lambda$ such that $f"= \lambda f$ and $g"= \lambda g$ when both f and g are linear combinations of $e^{\lamba x}$ and $e^{-\lambda x}$.

Last edited: Oct 14, 2009
3. Oct 13, 2009

### PhantomOort

Yes, both $$\lambda$$ and $$\mu$$ are numbers. Obviously this only works in cases where they exist, such as for exponentials, trig functions, or x. But these make up a great many common integrals, so it's a pretty good shortcut.

4. Oct 14, 2009

### trambolin

It does not seem to be a shortcut but rather like a nice exam question that my professors used to ask. I really liked it. Let me write it again, becaues latex parser went crazy above...

$$fg = (\lambda -\mu)\frac{fg}{\lambda -\mu} = \frac{f''g-fg''}{\lambda - \mu} = \frac{f''g-fg''+f'g' - f'g'}{\lambda - \mu} = \frac{(f'g)' -(fg')'}{\lambda - \mu}$$

Integration gives the result...

5. Oct 14, 2009

### HallsofIvy

Staff Emeritus
It would seem to me that integrals of the form $\int(Ae^{\lambda x}+ Be^{-\lamba x})(Ce^{\lambda x}+ De^{-\lambda x}) dx$$= \int (ACe^{2\lambda x}+ BDe^{-2\lambda x}+ (BC+ AD))dx$ could be done directly without worrying about integration by parts.

6. Oct 14, 2009

### PhantomOort

Damn, I could have done that! I need to stop being so lazy.

Thanks.

7. Oct 14, 2009

### PhantomOort

It's for integrals like $\int x sin(x) dx$, which are usually done by parts. Not that this is super hard, but with the shortcut, $$\lambda=0, \mu = -1$$, and the result is $$sin(x)-xcos(x)$$ and we're done.

PS, my apologies to any readers put off by the formatting. I'm utterly unfamiliar with this tex style and working through it as I go.