$\textbf{Problem:}$ I have the following improper integral in polar coordinates:
$$E =\int\limits_0^{2\pi} \int\limits_a^\infty I \cdot r \cdot dr \cdot d\theta$$
Rewriting the inner improper integral
$$E =\int\limits_0^{2\pi} \lim_{b\to\infty}\int\limits_a^b I \cdot r \cdot dr \cdot d\theta$$
$\textbf{Question: Is it then OK to interchange like this??}$
$$E =\lim_{b\to\infty} \int\limits_0^{2\pi} \int\limits_a^b I \cdot r \cdot dr \cdot d\theta$$
I ask because the latter converges to the result stated in my book, and the former blows up?…
I posted a similar question 5 months ago, but no one has answered it, so I decided to restate it a bit and try again 🙂
$\textbf{Background}$
I am trying to solve an integral I found in a book (Offshore Hydromechanics). This book states that one way to find the increase in kinetic energy of the potential flow around a cylinder $\textbf{(see attached pictures)}$ is by evaluating the following integral that subtracts the total initial energy from the total energy after the cylinder is inserted:
$$E =\iint\limits_{cylinder}^\infty \frac{1}{2}\cdot \rho \cdot [u(x,y,t)]^2 dx\cdot dy -\iint\limits_{cylinder}^\infty \frac{1}{2}\cdot \rho \cdot [u_\infty(t)]^2 dx\cdot dy $$
The book states it's source to be Hydrodynamics by Sir Horace Lamb. I found this book but was unable to locate this integral in the book after much searching. Then I decided to try and solve it myself.
I figured it was easier to solve in polar coordinates. This led me to the following vector fields:
The undisturbed vectorfield:
$$V = u_\infty \textbf{i} + 0\textbf{j}$$
And for the disturbed vectorfield in polar coordinates (a is cylinder radius):
$$V = \left(u_\infty \left(1-\frac{a^2}{r^2}\right)cos(\theta)\right)\textbf{v}_r + \left(- u_\infty\left(1+\frac{a^2}{r^2}\right)sin(\theta)\right)\textbf{v}_\theta $$
So the integral from above should be written (since u is the size of the velocity):
$$E =\int\limits_0^{2\pi} \int\limits_a^\infty \frac{1}{2}\cdot \rho \cdot \left[\sqrt{\left(u_\infty \left(1-\frac{a^2}{r^2}\right)cos(\theta)\right)^2+\left(- u_\infty\left(1+\frac{a^2}{r^2}\right)sin(\theta)\right)^2}\right]^2 r \cdot dr\cdot d\theta \\
-\int\limits_0^{2\pi} \int\limits_a^\infty \frac{1}{2}\cdot \rho \cdot u_\infty^2 r \cdot dr\cdot d\theta $$
I then moved the integrands under the same integral signs since the limits of integration were the same
$$E =\int\limits_0^{2\pi} \int\limits_a^\infty \left(\frac{1}{2}\rho \left[\left(u_\infty \left(1-\frac{a^2}{r^2}\right)cos(\theta)\right)^2+\left(- u_\infty\left(1+\frac{a^2}{r^2}\right)sin(\theta)\right)^2- u_\infty^2\right] \right) r dr d\theta $$
The inner integral is improper. If I evaluate the inner integral from a to b, then take the limit as b goes to infinity the entire integral does not converge.
However if I evaluate the inner integral from a to b, then evaluate the outer integral, and then afterwards take the limit as b goes to infinity, the integral converges nicely to the following:
$$E = \frac{1}{2}\cdot u_\infty^2\cdot a^2 \cdot\rho \cdot \pi $$
Which is equal to the kinetic energy of a mass equal to the mass of displaced water moving with the free flow speed (if integrating in z-coordinate). This is the correct answer according to my book…
Now my question is if this is allowed?? I mean interchanging the limit of the inner improper integral with the outer integral? I didn't think it was, but it gives me the correct answer? That's what made me wonder 🙂 Or are there other ways to solve the integral?
Best Answer
The integral can be written as
$$E = \frac{1}{2} \rho u_\infty^2 \int_0^{2\pi}\int_a^\infty \left[\frac{a^4}{r^4} - \frac{2 a^2}{r^2}(\cos^2 \theta - \sin^2 \theta)\right]r \, dr\, d\theta $$
Changing variables to $s = r/a$ and using the double angle formula for cosine we get
$$E = \frac{1}{2} \rho u_\infty^2 a^2\int_0^{2\pi}\int_1^\infty \left(\frac{1}{s^3} - \frac{2 }{s}\cos2 \theta \right) ds\, d\theta $$
As an iterated improper integral, the value depends on the order of integration, where
$$\pi = \int_1^\infty \int_0^{2\pi}\left(\frac{1}{s^3} - \frac{2 }{s}\cos2 \theta \right) d\theta\, ds \neq \int_0^{2\pi}\int_1^\infty \left(\frac{1}{s^3} - \frac{2 }{s}\cos2 \theta \right) ds\, d\theta,$$
since the RHS is divergent.
This is consistent with the fact that the integrand is not absolutely integrable over the infinite region with respect to the product measure and Fubini's theorem does not apply.
Clearly, regardless of the order of integration,
$$\int_0^{2\pi}\int_1^\infty \frac{1}{s^3} ds \,d\theta = \pi,$$
and so the problem arises due to the second term in the integrand, where by virtue of cancellation we have $\displaystyle\int_0^{2\pi} \frac{2}{s}\cos 2 \theta \, d \theta = 0$ , but $\displaystyle\int_1^\infty\frac{2}{s}\cos 2 \theta \, ds $ diverges for each $\theta$.
There are two issues here that lead to a problem in reconciling the mathematical result with physical intuition. One issue is that an unbounded domain cannot truly represent a real situation. Even uniform flow (the far-field condition) cannot exist in an unbounded domain as it requires infinite energy. The other is how a conditionally convergent improper integral is defined over an unbounded multi-dimensional domain in general.
A more (physically) agreeable approach is to consider a bounded domain where $(s,\theta) \in [1,b]\times [0,2\pi]$, and where $b$ can be very large but finite. We are interested in the energy of the disturbance flow which is given by the limit as $b \to \infty$ of
$$\frac{E_b}{\frac{1}{2} \rho u_\infty^2 a^2} = \int_1^b \int_0^{2\pi}\left(\frac{1}{s^3} - \frac{2 }{s}\cos2 \theta \right) d\theta\, ds = \int_0^{2\pi}\int_1^b \left(\frac{1}{s^3} - \frac{2 }{s}\cos2 \theta \right) ds\, d\theta$$
Now we can evaluate the iterated integrals in any order as the integrand is a continuous function on the bounded domain. It follows that
$$\frac{E_b}{\frac{1}{2} \rho u_\infty^2 a^2} = \pi\left(1 - \frac{1}{b^2}\right)- \log b \underbrace{\int_0^{2\pi} \cos 2\theta \, d\theta}_{=0} = \pi\left(1 - \frac{1}{b^2}\right),$$
and as $b \to \infty$
$$\frac{E_b}{\frac{1}{2} \rho u_\infty^2 a^2} \to \pi$$