Integrals III: Frullani, Beta, Gamma, Weierstrass & Series

Every integral of the form "difference of two scalings, divided by $x$" is Frullani: $\int_0^\infty\frac{e^{-x} - e^{-3x}}{x}dx = \ln 3$, $\int_0^\infty\frac{\cos x - \cos 2x}{x}dx = \ln 2$, $\int_0^\infty\frac{\arctan 2x - \arctan x}{x}dx = \tfrac\pi2\ln 2$. The $\tfrac1x$ makes each piece diverge alone; only the difference converges, and the value depends on $f$ only through its two endpoint values — which is why it looks like magic and is really Feynman's trick in disguise.

Gamma interpolates the factorial ($\Gamma(s+1) = s\Gamma(s)$ by one integration by parts) and evaluates every $\int_0^\infty x^ne^{-ax}dx = \tfrac{n!}{a^{n+1}}$. Beta evaluates every polynomial-times-polynomial integral on $[0,1]$ in one line ($\int_0^1x^3(1-x)^2dx = \tfrac{3!\,2!}{6!} = \tfrac1{60}$) and, via $x = \sin^2\theta$, every $\int_0^{\pi/2}\sin^m\theta\cos^n\theta\,d\theta = \tfrac12B(\tfrac{m+1}{2}, \tfrac{n+1}{2})$ — the Wallis integrals are Beta values. The reflection formula $\Gamma(s)\Gamma(1-s) = \tfrac{\pi}{\sin\pi s}$ is why $\int_0^\infty\frac{dx}{1+x^n} = \frac{\pi/n}{\sin(\pi/n)}$.

Weierstrass: $\sin x = \tfrac{2t}{1+t^2}$, $\cos x = \tfrac{1-t^2}{1+t^2}$, $dx = \tfrac{2\,dt}{1+t^2}$ — so $\int_0^\pi\frac{dx}{2 + \cos x} = \int_0^\infty\frac{2\,dt}{3 + t^2} = \frac{\pi}{\sqrt3}$. Series: $\int_0^1\frac{\ln(1+x)}{x}dx = \sum\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}$, and $\int_0^1\ln x\ln(1-x)dx = \sum\frac{1}{n(n+1)^2} = 2 - \frac{\pi^2}{6}$. The swap is legal under uniform convergence on the interval — say so.

Write $\dfrac{f(ax) - f(bx)}{x} = -\displaystyle\int_a^b f'(tx)\,dt$ (fundamental theorem in $t$; assume $f'$ continuous for the clean version). Then $\int_0^\infty\frac{f(ax) - f(bx)}{x}dx = -\int_a^b\left(\int_0^\infty f'(tx)\,dx\right)dt = -\int_a^b\frac{f(\infty) - f(0)}{t}dt = (f(0) - f(\infty))\ln\frac ba$, swapping the order of integration (Fubini, justified by absolute convergence). $\blacksquare$ Significance: the proof IS Feynman's trick — the $\tfrac1x$ was hiding an integral over a parameter. Once you see that, Frullani stops being a formula to memorize.

Substituting $x = u^2$ in $\Gamma(\tfrac12) = \int_0^\infty x^{-1/2}e^{-x}dx$ gives $2\int_0^\infty e^{-u^2}du = \int_{-\infty}^\infty e^{-u^2}du = I$. Square it and go polar: $I^2 = \iint e^{-(x^2+y^2)}dA = \int_0^{2\pi}\!\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\tfrac12 = \pi$. So $I = \sqrt\pi = \Gamma(\tfrac12)$. $\blacksquare$ No one-variable method evaluates this integral; the plane does. Every normal-distribution constant in statistics traces back to this square.

Step 1 — recognize $f(t) = \cos t$ with $a = 1$, $b = 2$; but $\cos t$ has no limit at $\infty$, so the plain theorem does not apply.

Step 2 — the fix: Frullani holds if $f(\infty)$ is replaced by the CESÀRO mean $\lim_{T\to\infty}\tfrac1T\int_0^Tf$, which for cosine is $0$.

Step 3 — the value is $(f(0) - 0)\ln\tfrac21 = \ln 2$.

Step 4 — verify with the damping trick: $\int_0^\infty e^{-sx}\frac{\cos x - \cos 2x}{x}dx = \tfrac12\ln\frac{s^2 + 4}{s^2 + 1} \to \ln 2$ as $s \to 0$ ✓. Knowing the theorem's exact hypotheses — and how to repair them — is the difference between a $10$ and a $1$.

Step 1 — $\ln(1 + x) = \sum_{n\ge1}\frac{(-1)^{n+1}x^n}{n}$ on $[0, 1]$ (convergent at $x = 1$ by the alternating series test; uniform on $[0,1]$ by Abel).

Step 2 — divide by $x$ and integrate term by term: $\sum_{n\ge1}\frac{(-1)^{n+1}}{n}\int_0^1x^{n-1}dx = \sum\frac{(-1)^{n+1}}{n^2}$.

Step 3 — this is $\eta(2) = (1 - 2^{1-2})\zeta(2) = \tfrac12\cdot\tfrac{\pi^2}{6} = \tfrac{\pi^2}{12}$.

Step 4 — numeric check: $1 - \tfrac14 + \tfrac19 - \tfrac1{16} + \cdots \approx 0.822$ and $\tfrac{\pi^2}{12} \approx 0.822$ ✓.

$\displaystyle\int_1^{\infty}\frac{dx}{x^p}$ converges iff $p > 1$; $\displaystyle\int_0^{1}\frac{dx}{x^p}$ converges iff $p < 1$. Those two facts, plus splitting, settle most convergence questions on sight: $\int_0^{\infty}x^{-p}\,dx$ diverges for EVERY $p$, because no single $p$ can satisfy both conditions at once. A Putnam answer that integrates through a singularity without splitting is marked down even when the number is right.

Limit comparison is the practical tool: near $\infty$, compare with $x^{-p}$; near a singularity at $a$, compare with $(x-a)^{-p}$. So $\int_1^\infty \frac{x + \sin x}{x^3 + 1}\,dx$ behaves like $\int x^{-2}$ and converges, while $\int_2^\infty \frac{dx}{x\ln x}$ behaves like a harmonic sum and diverges — substituting $u = \ln x$ makes that one exact.

For a symmetric singularity the CAUCHY PRINCIPAL VALUE takes the symmetric limit, $\mathrm{PV}\int_{-1}^{1}\frac{dx}{x} = 0$, but the integral itself does not converge — saying which one you are computing is part of the answer. Dirichlet-type integrals converge because of cancellation, which is why Feynman's trick and Dirichlet's test, not comparison, are the tools that handle them.

Step 1 — split at $1$: the integrand is improper at $0$ and at $\infty$.

Step 2 — near $0$, $\dfrac{1}{x^p(1+x)} \sim x^{-p}$, so the piece on $(0,1]$ converges iff $p < 1$.

Step 3 — near $\infty$, $\dfrac{1}{x^p(1+x)} \sim x^{-p-1}$, so the piece on $[1,\infty)$ converges iff $p + 1 > 1$, i.e. $p > 0$.

Step 4 — both together: $0 < p < 1$. (For those $p$ the value is $\dfrac{\pi}{\sin \pi p}$, the Beta-function evaluation.)

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

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

Fresnel is the Mellin integral $\int_0^\infty t^{s-1}\sin t\,dt = \Gamma(s)\sin\frac{\pi s}{2}$ at $s = \frac12$. The reflection formula is why $\int_0^\infty\frac{x^{s-1}}{1 + x}dx = \frac{\pi}{\sin\pi s}$ and hence $\int_0^\infty\frac{dx}{1 + x^n} = \frac{\pi/n}{\sin(\pi/n)}$. Wallis comes from the sine-power reduction formula squeezed between consecutive terms; $\int_0^1\frac{\ln x}{1-x}$ is the series method landing on $\zeta(2)$.

Let $u = \sin^2x$, $dv = x^{-2}dx$, $v = -x^{-1}$: $\int_0^\infty\frac{\sin^2x}{x^2}dx = \left[-\frac{\sin^2x}{x}\right]_0^\infty + \int_0^\infty\frac{2\sin x\cos x}{x}dx$. The boundary term is $0$ at both ends ($\sin^2x \sim x^2$ near $0$; bounded over $x$ at infinity). The remaining integrand is $\frac{\sin 2x}{x}$; substituting $t = 2x$ gives $\int_0^\infty\frac{\sin t}{t}dt = \frac\pi2$. $\blacksquare$ Significance: the equality of $\int\frac{\sin x}{x}$ and $\int\frac{\sin^2 x}{x^2}$ is not a coincidence but a one-line consequence of parts — and the pattern extends to the Fourier transform of a triangle function.

$\Gamma(s)\Gamma(1-s) = \Gamma(1)B(s, 1-s) = \int_0^1t^{s-1}(1-t)^{-s}dt$; substitute $t = \frac{x}{1+x}$ to get $\int_0^\infty\frac{x^{s-1}}{1+x}dx$. This last integral equals $\frac{\pi}{\sin\pi s}$ — provable without complex analysis by expanding $\frac{1}{1+x}$ as a geometric series on $(0,1)$ and using $x \to \frac1x$ on $(1, \infty)$, which yields $\sum_{k\ge0}(-1)^k\left(\frac{1}{s+k} + \frac{1}{1 - s + k}\right)$, the partial-fraction expansion of $\frac{\pi}{\sin\pi s}$. $\blacksquare$ Significance: $s = \frac12$ gives $\Gamma(\frac12)^2 = \pi$ — the Gaussian integral yet again — and the formula evaluates every $\int_0^\infty\frac{dx}{1 + x^n}$.