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Thread: Integration by parts problem?!?!

  1. #1
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    Exclamation 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
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  2. #2
    MHF Contributor

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

    Of course however, since $\displaystyle e^xf(x)=e^{x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)$, two integrations by parts would suffice if you split $\displaystyle e^xf(x)$ into $\displaystyle u(x)=e^{x/2}$ and $\displaystyle v'(x)=\cos\left(\frac{\sqrt{3}}{2}x\right)$. 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.
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  3. #3
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    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.
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