Calculus in casinos

The question implicitly asks for the expected value of payout. This is equal to $\displaystyle \frac{1}{1-0} \int_{[0,1]} \frac{dx}{x^3+1}$ . The integral can evaluated with partial fractions like so:

$\begin{aligned} \frac{1}{1-0} \int_{[0,1]} \frac{dx}{x^3+1} &= \int_0^1 \frac{2-x}{3(x^2-x+1)}+\frac{1}{3(x+1)} \ dx \\ &= \frac{1}{3} \int_0^1 \frac{2-x}{x^2-x+1} \ dx + \frac{1}{3} \int_0^1 \frac{dx}{x+1} \\ &= \frac{1}{3} \int_0^1 \frac{2-x}{x^2-x+1} \ dx + \frac{1}{3}[\ln(1+1) - \ln(0+1)] \\ &= \frac{1}{3} \int_0^1 \frac{2-x}{x^2-x+1} \ dx + \frac{\ln 2}{3} \end{aligned}$

We wish to turn the denominator of the first integrand into some quadratic of the form $u^2+1$ , where $u$ is a linear function of $x$ , to invoke a trigonometric substitution. Solving $(ax+b)^2+1 = a^2(x^2-x+1)$ (with the factor of $x^2-x+1$ necessarily $a^2$ ), we get $u = \frac{2x-1}{\sqrt 3}$ as a solution. Thus, using u-substitution, we get

$\int_0^1 \frac{2-x}{x^2-x+1} \ dx = \int_{-1/\sqrt 3}^{1/\sqrt 3} \frac{2-(\sqrt 3u+1)/2}{3(u^2+1)/4} \ \frac{du}{2/\sqrt 3} = \int_{-1/\sqrt 3}^{1/\sqrt 3} \frac{\sqrt 3-u}{u^2+1} \ du$

We are now prepared to invoke the trigonometric substitution; namely, we let $\theta = \tan^{-1} u$ , and thus,

$\int_{-1/\sqrt 3}^{1/\sqrt 3} \frac{\sqrt 3-u}{u^2+1} \ du = \int_{-\pi/6}^{\pi/6} \frac{\sqrt 3-\tan \theta}{u^2+1} \ \frac{d\theta}{1/(u^2+1)} = \int_{-\pi/6}^{\pi/6} \sqrt 3-\tan \theta \ d\theta$

Note that $\tan$ is an odd function, and so

$\int_{-\pi/6}^{\pi/6} \sqrt 3-\tan \theta \ d\theta = \sqrt 3 \int_{-\pi/6}^{\pi/6} \ d\theta = \sqrt 3 \left[ \frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right] = \frac{\sqrt 3 \pi}{3}$

Putting it altogether, we finally have

$\begin{aligned} \frac{1}{1-0} \int_{[0,1]} \frac{1}{x^3+1} \ dx &= \frac{1}{3} \int_0^1 \frac{2-x}{x^2-x+1} \ dx + \frac{1}{3} \int_0^1 \frac{dx}{x+1} \\ &= \frac{1}{3} \frac{\sqrt 3 \pi}{3} + \frac{\ln 2}{3} \\ & \approx \boxed{0.8356} \end{aligned}$