# Bizarre integral solved using conservation of energy

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.

• Also, if you have a solution via conservation of energy then it would seem worth including as its own answer. Commented Jun 16 at 2:16

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 $$I=2+\frac{3}{2\beta}$$.

Following Semiclassical's answer, the second integral is not difficult as well. $$\int_0^1\sqrt{2y-y^2}df(y)=\sqrt{2y-y^2}f(y)\bigg|_0^1-\int_0^1f(y)d\left(\sqrt{2y-y^2}\right)$$ $$=2-\int_0^1\frac{(3-y)\sqrt{y}}{(2-y)^{3/2}}\frac{2-2y}{2\sqrt{2y-y^2}}dy$$ $$=2-\int_0^1\frac{(3-y)(1-y)}{(2-y)^2}dy$$ $$=2-\int_1^2\frac{(u+1)(u-1)}{u^2}du$$ $$=2-\int_1^2\left(1-\frac{1}{u^2}\right)du$$ $$=2-\left(u+\frac{1}{u}\right)\bigg|_1^2$$ $$=2-1+\frac{1}{2}$$ $$=\frac{3}{2}$$