# Integration by parts problem?!?!

• Sep 22nd 2008, 10:49 AM
t_n83
Integration by parts problem?!?!
I know how to do integration by parts. ∫ u dv= uv-∫ v du

as you may know for some functions like ∫ e^x cosx dx when you use integration by parts it repeats it's self. if you don't know what I mean, see the 3rd example on this page http://www.math.hmc.edu/calculus/tutoria...

what I need to know is a function that repeats when you use itegration by parts like ∫ e^x cosx dx
but instead of 2 steps it has 3.

if anyone knows of one let me know
• Sep 22nd 2008, 01:08 PM
Laurent
The following example is very artificial and therefore not quite satisfying, but it answers you question anyway.
Consider $f(x)=e^{-\frac{x}{2}}\cos\left(\frac{\sqrt{3}}{2}x\right)$. You can check that $f'''(x)=f(x)$, so that you may compute $\int f(x)e^x\,dx$ using 3 integrations by parts, which will end up with the initial integral.
In fact, if you know complex numbers, $f(x)$ is the real part of $e^{jx}$ where $j=e^{\frac{2i\pi}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ is a cubic root of the unity: $j^3=1$, so what I claimed about $f$ is a simple consequence of this property.

Of course however, since $e^xf(x)=e^{x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)$, two integrations by parts would suffice if you split $e^xf(x)$ into $u(x)=e^{x/2}$ and $v'(x)=\cos\left(\frac{\sqrt{3}}{2}x\right)$. I don't know of a genuine example where after 3 integrations by parts, and not possibly less, we recover the initial integral, in a way enabling to find its value.
• Sep 22nd 2008, 01:43 PM
t_n83
Thanks, Laurent.

If anyone can find a specific example of, as Laurent put it, where after 3 integrations by parts, and not possibly less, we recover the initial integral, that'd be great.