# Integration by Parts and Series

1. Jun 4, 2014

### Husaaved

This isn't really a homework question, more just something I noticed while evaluating an integral and was curious about:

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 coefficientâ€‹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. Jun 4, 2014

### jbunniii

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
3. Jun 4, 2014

### Husaaved

Thank you so much, I wasn't expecting such a detailed or helpful response.

4. Jun 4, 2014

### lurflurf

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.