Integrals IV: Dirichlet Variants, Fresnel, Wallis & the Reflection Formula

Study Sheet

Integrals IV: Dirichlet Variants, Fresnel, Wallis & the Reflection Formula

The second shelf of named integrals, each with its proof

More Named Results

Tip
Dirichlet variants by parts

∫0∞sin⁡2xx2dx=∫0∞1−cos⁡xx2dx=π2\int_0^\infty\frac{\sin^2x}{x^2}dx = \int_0^\infty\frac{1 - \cos x}{x^2}dx = \frac\pi2 — one integration by parts each, landing on ∫0∞sin⁡xxdx\int_0^\infty\frac{\sin x}{x}dx.

Reminder — Substitution and integration by parts:  ∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du

With u=sin⁡2xu = \sin^2x and dv=x−2dxdv = x^{-2}dx the boundary term vanishes and the remainder is ∫sin⁡2xxdx\int\frac{\sin 2x}{x}dx, which rescales to Dirichlet. The same move on 1−cos⁡x1 - \cos x gives Dirichlet directly. A whole family — ∫0∞(sin⁡xx)ndx\int_0^\infty\left(\frac{\sin x}{x}\right)^n dx — follows by repeated parts and is a Putnam regular for n=3n = 3 (3π8\frac{3\pi}{8}).

Tip
Log-trig integrals and the tangent bridge

∫0π/2ln⁡cos⁡x dx=−π2ln⁡2\int_0^{\pi/2}\ln\cos x\,dx = -\frac\pi2\ln 2, hence ∫0∞ln⁡(1+x2)1+x2dx=πln⁡2\int_0^\infty\frac{\ln(1 + x^2)}{1 + x^2}dx = \pi\ln 2 (substitute x=tan⁡θx = \tan\theta) and ∫0π/2ln⁡tan⁡x dx=0\int_0^{\pi/2}\ln\tan x\,dx = 0.

The tangent substitution converts every rational-times-log integrand on (0,∞)(0, \infty) into a log-trig integral on (0,π2)(0, \frac\pi2), where King and Queen apply. Feynman handles the arctangent cousin: ∫0∞arctan⁡xx(1+x2)dx=π2ln⁡2\int_0^\infty\frac{\arctan x}{x(1 + x^2)}dx = \frac\pi2\ln 2 via the parameter arctan⁡(ax)\arctan(ax).

Tip
Fresnel, the reflection formula, and Wallis

∫0∞sin⁡(x2)dx=∫0∞cos⁡(x2)dx=12π2\int_0^\infty\sin(x^2)dx = \int_0^\infty\cos(x^2)dx = \frac12\sqrt{\frac\pi2}; Γ(s)Γ(1−s)=πsin⁡πs\Gamma(s)\Gamma(1 - s) = \frac{\pi}{\sin\pi s}; ∏n≥14n24n2−1=π2\prod_{n\ge1}\frac{4n^2}{4n^2 - 1} = \frac\pi2; ∫01ln⁡x1−xdx=−π26\int_0^1\frac{\ln x}{1 - x}dx = -\frac{\pi^2}{6}.

Fresnel is the Mellin integral ∫0∞ts−1sin⁡t dt=Γ(s)sin⁡πs2\int_0^\infty t^{s-1}\sin t\,dt = \Gamma(s)\sin\frac{\pi s}{2} at s=12s = \frac12. The reflection formula is why ∫0∞xs−11+xdx=πsin⁡πs\int_0^\infty\frac{x^{s-1}}{1 + x}dx = \frac{\pi}{\sin\pi s} and hence ∫0∞dx1+xn=π/nsin⁡(π/n)\int_0^\infty\frac{dx}{1 + x^n} = \frac{\pi/n}{\sin(\pi/n)}. Wallis comes from the sine-power reduction formula squeezed between consecutive terms; ∫01ln⁡x1−x\int_0^1\frac{\ln x}{1-x} is the series method landing on ζ(2)\zeta(2).

Side note
A note on rigor

Every result on this shelf involves an interchange — of limit and integral, of sum and integral, of the order of a double integral. On the Putnam, one sentence naming the justification (dominated convergence with a named dominating function, uniform convergence on the interval, absolute convergence for Fubini) is the difference between a 1010 and a 77. Say it every time.

Try it
Try it: from Dirichlet to its cube

Given ∫0∞sin⁡xxdx=π2\int_0^\infty\frac{\sin x}{x}dx = \frac\pi2, evaluate ∫0∞sin⁡3xx3dx\int_0^\infty\frac{\sin^3x}{x^3}dx.

Use sin⁡3x=3sin⁡x−sin⁡3x4\sin^3x = \frac{3\sin x - \sin 3x}{4} and integrate by parts twice to reduce ∫sin⁡3xx3\int\frac{\sin^3x}{x^3} to ∫sin⁡x−(rescaled)x\int\frac{\sin x - \text{(rescaled)}}{x} terms; the result is 3π8\frac{3\pi}{8}.

Proofs & Why It Matters

Tip
Proof: ∫0∞sin⁡2xx2dx=π2\int_0^\infty\frac{\sin^2x}{x^2}dx = \frac\pi2

Parts onto Dirichlet.

Let u=sin⁡2xu = \sin^2x, dv=x−2dxdv = x^{-2}dx, v=−x−1v = -x^{-1}: ∫0∞sin⁡2xx2dx=[−sin⁡2xx]0∞+∫0∞2sin⁡xcos⁡xxdx\int_0^\infty\frac{\sin^2x}{x^2}dx = \left[-\frac{\sin^2x}{x}\right]_0^\infty + \int_0^\infty\frac{2\sin x\cos x}{x}dx. The boundary term is 00 at both ends (sin⁡2x∼x2\sin^2x \sim x^2 near 00; bounded over xx at infinity). The remaining integrand is sin⁡2xx\frac{\sin 2x}{x}; substituting t=2xt = 2x gives ∫0∞sin⁡ttdt=π2\int_0^\infty\frac{\sin t}{t}dt = \frac\pi2. ■\blacksquare Significance: the equality of ∫sin⁡xx\int\frac{\sin x}{x} and ∫sin⁡2xx2\int\frac{\sin^2 x}{x^2} is not a coincidence but a one-line consequence of parts — and the pattern extends to the Fourier transform of a triangle function.

Tip
Proof: the reflection formula Γ(s)Γ(1−s)=πsin⁡πs\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin\pi s} for 0<s<10 < s < 1

Via the Beta function and a contour-free evaluation.

Γ(s)Γ(1−s)=Γ(1)B(s,1−s)=∫01ts−1(1−t)−sdt\Gamma(s)\Gamma(1-s) = \Gamma(1)B(s, 1-s) = \int_0^1t^{s-1}(1-t)^{-s}dt; substitute t=x1+xt = \frac{x}{1+x} to get ∫0∞xs−11+xdx\int_0^\infty\frac{x^{s-1}}{1+x}dx. This last integral equals πsin⁡πs\frac{\pi}{\sin\pi s} — provable without complex analysis by expanding 11+x\frac{1}{1+x} as a geometric series on (0,1)(0,1) and using x→1xx \to \frac1x on (1,∞)(1, \infty), which yields ∑k≥0(−1)k(1s+k+11−s+k)\sum_{k\ge0}(-1)^k\left(\frac{1}{s+k} + \frac{1}{1 - s + k}\right), the partial-fraction expansion of πsin⁡πs\frac{\pi}{\sin\pi s}. ■\blacksquare Significance: s=12s = \frac12 gives Γ(12)2=π\Gamma(\frac12)^2 = \pi — the Gaussian integral yet again — and the formula evaluates every ∫0∞dx1+xn\int_0^\infty\frac{dx}{1 + x^n}.