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Integration by Parts and Series

  1. Jun 4, 2014 #1
    This isn't really a homework question, more just something I noticed while evaluating an integral and was curious about:

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    At this stage, I was able to simplify the expression before solving for the integral algebraically (since the second iteration yielded the original integral the right side) and in doing so I came to the correct answer. I noticed however that I could have kept applying integration by parts over and over again.

    Had I done this, would it be possible to evaluate the integral by applying some property of series to the coefficient​s to see if they converge at some value? I know that the best way to solve the integral is the method I ended up using, but I was just curious if there were some relationship between integrals, coefficients, and series. I only have a conceptual understanding of series at this point so pardon me if my question isn't a valid one.

    Thank you.
     
    Last edited: Jun 4, 2014
  2. jcsd
  3. Jun 4, 2014 #2

    jbunniii

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    I'm not sure exactly what you're looking for, but we can get some insight into this problem using linear algebra.

    Throughout, I will ignore the "##+C##" associated with the indefinite integral, or equivalently, choose ##C = 0##.

    If we let ##f(\theta) = e^{2\theta}\sin(3\theta)## and ##g(\theta) = e^{2\theta}\cos(3\theta)##, then we can consider ##f## and ##g## as elements of the vector space of all continuous functions. Clearly ##f## and ##g## are linearly independent: one is not a scalar multiple of the other. So ##f## and ##g## span a subspace of dimension 2. Note that integration is a linear operator on this vector space. If we denote this operator by ##L##, your calculations show that
    $$L(f) = -\frac{1}{3} g + \frac{2}{3}L(g)$$
    $$L(g) = \frac{1}{3}f - \frac{2}{3}L(f)$$
    We may express ##L## as a matrix ##M## with respect to the basis ##\{f,g\}## as follows:
    $$M\begin{pmatrix}1 \\ -2/3 \end{pmatrix} = \begin{pmatrix}0 \\ -1/3 \end{pmatrix}$$
    $$M\begin{pmatrix}2/3 \\ 1 \end{pmatrix} = \begin{pmatrix}1/3 \\ 0 \end{pmatrix}$$
    Stacking these together into one equation, we get
    $$M\begin{pmatrix}1 & 2/3 \\ -2/3 & 1 \end{pmatrix} = \begin{pmatrix}0 & 1/3 \\ -1/3 & 0\end{pmatrix}$$
    Solving for ##M## gives us
    $$M = \begin{pmatrix}2/13 & 3/13 \\ -3/13 & 2/13 \end{pmatrix} = \frac{1}{\sqrt{13}} \begin{pmatrix}2/\sqrt{13} & 3/\sqrt{13} \\ -3/\sqrt{13} & 2/\sqrt{13} \end{pmatrix} $$
    Since ##(2/\sqrt{13})^2 + (3/\sqrt{13})^2 = 1##, we see that ##M## is of the form
    $$\frac{1}{\sqrt{13}}\begin{pmatrix}\cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{pmatrix}$$
    which is a rotation by the angle ##x = \cos^{-1}(2/\sqrt{13})## and a scaling by ##1/\sqrt{13}##.

    This makes it quite easy to see what the answer will be if you apply the operator (i.e. integrate) repeatedly, say ##n## times. Since
    $$M^n = \left(\frac{1}{\sqrt{13}}\right)^n \begin{pmatrix}\cos(nx) & \sin(nx) \\ -\sin(nx) & \cos(nx) \end{pmatrix}$$
    the result of integrating your ##f## function ##n## times will be
    $$\begin{align}
    L^n(f) &=
    M^n\begin{pmatrix}1 \\ 0\end{pmatrix} \\
    &= \left(\frac{1}{\sqrt{13}}\right)^n \begin{pmatrix}\cos(nx) \\ -\sin(nx)\end{pmatrix} \\
    &= \left(\frac{1}{\sqrt{13}}\right)^n (f(\theta) \cos(nx) -g(\theta)\sin(nx))\end{align}$$
     
    Last edited: Jun 4, 2014
  4. Jun 4, 2014 #3
    Thank you so much, I wasn't expecting such a detailed or helpful response.
     
  5. Jun 4, 2014 #4

    lurflurf

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    I think
    jbunniii interpreted the question a little differently than I did.

    If we integrate by parts 2n times we will see

    $$\int \! e^{2\theta}\sin(3\theta) \, \mathrm{d}\theta=
    \tfrac{1}{13}e^{2\theta}[2\sin(3\theta)-3\cos(3\theta)]+ \\
    \left(-\tfrac{4}{9}\right)^n \left\{ -\tfrac{1}{13}e^{2\theta}[2\sin(3\theta)
    -3\cos(3\theta)]+\int \! e^{2\theta}\sin(3\theta) \, \mathrm{d}\theta \right\}$$

    from which we make the same conclusion as before.
     
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