Name for integration by parts shortcut

PhantomOort
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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 \intf(x)g(x)dx = \frac{f'g-fg'}{\lambda-\mu}.

Can anyone tell me what this is called? Thanks.
 
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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}.
 
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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.
 
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...
 
PhantomOort said:
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.
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.
 
trambolin said:
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...
Damn, I could have done that! I need to stop being so lazy.

Thanks.
 
HallsofIvy said:
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.

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.
 
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