Integrals II: King's Rule, Queen's Rule, and Glasser

The substitution $u = a + b - x$ runs the interval backward with $du = -dx$; flipping the limits back absorbs the sign. Nothing is evaluated yet — the power comes from adding the reflected copy to the original: $2I = \int_a^b\left[f(x) + f(a+b-x)\right]dx$, and the bracket often collapses. On $[0, \tfrac\pi2]$, sine and cosine trade places; on $[0, \pi]$, $\sin$ is fixed and $\cos$ flips sign; a stray factor $x$ becomes $a + b - x$, so $x + (a + b - x) = a + b$ is constant. Model: $I = \int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}dx$; reflected, $\frac{\cos x}{\cos x + \sin x}$; sum $= 1$; $I = \tfrac\pi4$.

Both are King's rule applied to half the interval: split $[0, 2a]$ at $a$, reflect the right half onto the left, and compare. Use them to FOLD an interval before doing any real work: $\int_0^\pi\sin^3x\,dx = 2\int_0^{\pi/2}\sin^3x\,dx = \tfrac43$, and any odd integrand over a symmetric interval is $0$ on sight ($\int_{-1}^1 x^3e^{x^2}dx = 0$ — no computation). Half the Putnam's "obviously zero" integrals are Jack's rule.

The map $u = x - \tfrac1x$ is increasing on $(-\infty, 0)$ and on $(0, \infty)$, covering all of $\mathbb R$ on each half-line. Splitting the integral at $0$ and substituting on each half, the two Jacobian pieces $\tfrac{dx}{du}$ ADD to exactly $1$ — the identity $\sum_{\text{branches}}\tfrac{dx}{du} = 1$ is the content of the theorem. So $\int_{-\infty}^\infty e^{-(x - 1/x)^2}dx = \int e^{-u^2}du = \sqrt\pi$: a hideous integrand, evaluated by recognizing its shape. The companion on $(0, \infty)$ is $x \to \tfrac1x$ alone, which maps the half-line to itself.

Reflect: $I = \int_0^\pi\frac{(\pi - x)\sin x}{1 + \cos^2x}dx$. Add: $2I = \pi\int_0^\pi\frac{\sin x}{1 + \cos^2x}dx$. Substitute $u = \cos x$: $\pi\int_{-1}^1\frac{du}{1 + u^2} = \pi\cdot\tfrac\pi2$. So $I = \tfrac{\pi^2}{4}$.

On $x > 0$ set $u = x - \tfrac1x$; solving, $x = \tfrac{u + \sqrt{u^2 + 4}}{2}$, an increasing bijection $(0,\infty) \to \mathbb R$ with $\tfrac{dx}{du} = \tfrac12 + \tfrac{u}{2\sqrt{u^2+4}}$. On $x < 0$ the same $u$ gives $x = \tfrac{u - \sqrt{u^2+4}}{2}$, also a bijection onto $\mathbb R$, with $\tfrac{dx}{du} = \tfrac12 - \tfrac{u}{2\sqrt{u^2+4}}$. Adding the two contributions to the integral: $\int f(u)\left[\tfrac12 + \tfrac{u}{2\sqrt{u^2+4}} + \tfrac12 - \tfrac{u}{2\sqrt{u^2+4}}\right]du = \int f(u)\,du$. $\blacksquare$ Significance: the general theorem with $\sum\tfrac{a_i}{x - b_i}$ works identically — the $n + 1$ branches' Jacobians always sum to $1$ (a partial-fractions identity). It is a favorite because the integrand looks impossible and the solution is three lines.

King gives $I = \int_0^{\pi/2}\ln(\cos x)dx$. Adding: $2I = \int_0^{\pi/2}\ln(\sin x\cos x)dx = \int_0^{\pi/2}\ln\tfrac{\sin 2x}{2}dx = \int_0^{\pi/2}\ln(\sin 2x)dx - \tfrac\pi2\ln 2$. In the remaining integral substitute $u = 2x$: $\tfrac12\int_0^\pi\ln(\sin u)du$, and Queen's rule folds $[0, \pi]$ to give $\tfrac12\cdot2I = I$. Hence $2I = I - \tfrac\pi2\ln 2$ and $I = -\tfrac\pi2\ln 2$. $\blacksquare$ The integral was never computed; it was trapped in an equation with itself.

Step 1 — substitute $x \to \tfrac1x$: $I = \int_0^\infty\dfrac{x^2\,dx}{1 + x^4}$.

Step 2 — add: $2I = \int_0^\infty\dfrac{1 + x^2}{1 + x^4}dx$.

Step 3 — divide numerator and denominator by $x^2$: $\dfrac{1 + x^{-2}}{x^2 + x^{-2}} = \dfrac{1 + x^{-2}}{(x - x^{-1})^2 + 2}$.

Step 4 — with $u = x - \tfrac1x$ (so $du = (1 + x^{-2})dx$ and $u$ runs over $\mathbb R$): $2I = \int_{-\infty}^\infty\dfrac{du}{u^2 + 2} = \dfrac{\pi}{\sqrt2}$.

Step 5 — $I = \dfrac{\pi}{2\sqrt2} = \dfrac{\pi\sqrt2}{4}$. Cross-check with the general formula $\int_0^\infty\frac{dx}{1 + x^n} = \frac{\pi/n}{\sin(\pi/n)}$ at $n = 4$: $\frac{\pi/4}{\sqrt2/2} = \frac{\pi\sqrt2}{4}$ ✓.

Step 1 — substitute $x = \tan\theta$: $I = \int_0^{\pi/4}\ln(1 + \tan\theta)\,d\theta$.

Step 2 — King on $[0, \tfrac\pi4]$: $\theta \to \tfrac\pi4 - \theta$, and $1 + \tan(\tfrac\pi4 - \theta) = 1 + \dfrac{1 - \tan\theta}{1 + \tan\theta} = \dfrac{2}{1 + \tan\theta}$.

Step 3 — so $I = \int_0^{\pi/4}\left[\ln 2 - \ln(1 + \tan\theta)\right]d\theta = \tfrac\pi4\ln 2 - I$.

Step 4 — $I = \dfrac{\pi\ln 2}{8}$. Two substitutions, no antiderivative — and the answer, $\tfrac{\pi}{8}\ln 2 \approx 0.272$, is one of the most famous values in the subject.