1

Motivated by the following thread What is integration by parts, really?, I started looking at the integral of $e^{x}\cos(x)$ by noting that $$\frac{d}{dx} e^{ax}\cos(bx) = ae^{ax}\cos(bx) -e^{ax}\sin(bx).$$ However, instead of using the standard integration by parts trick to solve this problem, you could instead write $$\frac{d}{dx} e^{ax} \sin(bx) = be^{ax}\cos(bx)+ ae^{ax}\sin(bx) $$ and use these two equations to find the matrix representation of the differentiation operator with respect to the basis $\{e^{ax}\cos(bx),e^{ax}\sin(bx)\}$. To me, this seems wonderful, the product rule or integration by parts seems to allow you to decompose a complicated function into linear pieces with respect to a basis.

Using this seems to let you integrate even more complicated functions like $e^{af(x)}\cos(bf(x))$ since $f$ only changes your coordinates and not the matrix representation of the operator.

I was able to solve a few other integrals, and I turned my attention to what would happen if I tried to use this technique on Fresnel integrals, the error function, $\sin(x)/x$ and other integrals with no closed form solution in terms of elementary functions, but everything seems to fail because I need a non-finite basis.

At this point, I have a few questions:

1) Is this really just integration by parts in disguise? I ask this because the product rule looks like it is just helping me find an ansatz basis to use for my inverse transformation.

2) Are there any interesting well known integrals that be solved more easily doing this instead of more common integration techniques?

3) If we are given an elementary function as the integrand, it kind of looks like I can get a hint as to whether or not its integral is also an elementary function by examining the size of the basis I would need to write down inverse operator of the derivative. Intuitively, I don't think the dimension should tell me anything, does it? I feel like this should be an obvious corollary of Taylor's theorem, but I cannot see it.

Community
  • 1
JessicaK
  • 7,655
  • 1
    "Using this seems to let you integrate even more complicated functions like $e^{af(x)}\cos(bf(x))$." Are you sure? What about the integral of $e^{ax^2}\cos(bx^2)$? (I'm genuinely asking, not dropping a hint - I assume this integral has no closed form but maybe I am wrong.) –  Jun 16 '14 at 00:32
  • @NotNotLogical I am not 100% sure, but if you do the calculation, it seems fine? The matrix representation would be

    $$T = \begin{bmatrix} a &b\ -b & a \end{bmatrix}$$

    with respect to the basis $B = {2xe^{ax^{2}}\cos(bx^{2}), 2xe^{ax^{2}}\sin(bx^{2})}$. Of course, $a$ and $b$ need to be given so that the matrix is non-singular.

     – JessicaK Jun 16 '14 at 00:38
  • I think there's a mistake somewhere: wolfram says that the integral has no elementary form: http://integrals.wolfram.com/index.jsp?expr=e%5E%28ax%5E2%29cos%28bx%5E2%29&random=false –  Jun 16 '14 at 00:40
  • @NotNotLogical I think I see the problem: When I differentiate your function, the added $x$ term is causing a problem because the basis I just wrote down is going to be linearly independent of the function I want to integrate when $f(x) = x^{2}$. I think it may be best to leave my question unedited because they might highlight some of my misunderstandings? This was the same issue I was having trying to apply this to the Fresnel Integrals – JessicaK Jun 16 '14 at 00:47

0 Answers0