I was working through a problem I invented while devising homework problems for my thermodynamics course.
I encountered a crazy integral and was able to show using conservation of energy that the integral has a very simple solution.
I checked the answer using numerical integration and it holds to arbitrary accuracy.
The integral $I$ is
$$
I = \int_1^{x'} \frac{df(x)}{x}
$$
where
$$
x' = \frac{\beta}{\beta+1}
$$
with $\beta > 0$ and
$$
f(x)=\frac{(3-y)y^{1/2}}{(2-y)^{3/2}}
$$
where
$$
y(x)=1 – \sqrt{1 – \left(\beta \times \frac{1-x}{x}\right)^2}
$$
I find that
$$
I = 2 + \frac{3}{2} \times \frac{1}{\beta}
$$
How can such a complicated looking integral have such a simple solution? The problem is essentially a calculation of the work required to inflate a soap bubble.
Best Answer
This is not a complete solution---see velut luna's answer for the calculation of one of the definite integrals---but explains the behavior seen.
Note that $x\mapsto y(x)$ maps $[\beta/(\beta+1),1]$ to $[0,1]$, so the integral can be expressed in terms of $y$: $$I=\int_0^1 \frac{df(y)}{x(y)}$$ Moreover, it's not too bad to solve for $x$ in terms of $y$:
\begin{align} y&=1-\sqrt{1-\left(\beta\frac{1-x}{x}\right)^2}\\ 1-(1-y)^2&=\left(\beta\frac{1-x}{x}\right)^2\\ 1-\frac{1}{x}&=-\beta^{-1}\sqrt{1-(1-y)^2}\\ \implies x(y)&=\left[1+\beta^{-1}\sqrt{2y-y^2}\right]^{-1} \end{align} where the minus sign of the square root is taken to recover $x(1)=\beta/(\beta+1)$. Hence the original integral can be expressed as
$$I=\int_0^1 \left(1+\beta^{-1}\sqrt{2y-y^2}\right)~df(y)$$
This is now evidently of the form $I=I_0+\beta^{-1} I_1$ for appropriate $\beta$-independent definite integrals $I_0,I_1$, verifying the functional form observed by the OP. The first can be immediately integrated to obtain $I_0=\int_0^1 df(y)=f(1)-f(0)=2.$ The second is not so obvious and I don't have a satisfactory solution. But according to Mathematica the integrand can be expressed as
$$\sqrt{2y-y^2}\frac{df}{dy}=\frac{d}{dy}\left(\frac{3}{2-y}\right)$$ and hence $I_2=3/(2-1)-3/(2-0)=3/2$. This verifies $I=2+\frac{3}{2\beta}$.