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 $b \to \infty$
$$\frac{E_b}{\frac{1}{2} \rho u_\infty^2 a^2} \to \pi$$
Best Answer
Shifting the argument of the integrand by $\pi$ amounts to multiplying by $-e^{-\pi}$.
Assume you know that
$$J=\int_0^\infty e^{-x}\sin x\, dx=\frac12,$$ which is easily established with $e^{-x}\sin x=\Im{e^{(-1+i)x}}\to J=-\Im(-1+i)^{-1}$.
Then
$$K=J+Je^{-\pi}=\frac12(1+e^{-\pi})$$ is the integral of the first arch (i.e. from $0$ to $\pi$), by cancellation of the other arches. And by summing on all rectified arches (that form a geometric series), the requested integral is
$$I=\frac K{1-e^{-\pi}}=\frac12\frac{1+e^{-\pi}}{1-e^{-\pi}}=\frac12\coth\frac\pi2.$$