Non-Elementary Integration

Embed the integral in a family $I(a) = \int f(x, a)\,dx$, compute $I'(a) = \int \partial_a f\,dx$ (usually elementary), integrate back in $a$, and fix the constant from a value of $a$ where $I$ is known. Example: $I(a) = \displaystyle\int_0^1\frac{x^a - 1}{\ln x}dx$ has $I'(a) = \int_0^1 x^a dx = \tfrac{1}{a+1}$, so $I(a) = \ln(a+1)$ (using $I(0) = 0$). The $\ln x$ that made the integral impossible was erased by one $\partial_a$.

$\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$ — substituting $x \to a + b - x$ flips the interval onto itself. Its power: ADD the two forms. For $I = \int_0^{\pi/2}\tfrac{\sin x}{\sin x + \cos x}dx$, the reflection swaps $\sin$ and $\cos$, and adding gives $2I = \int_0^{\pi/2}1\,dx = \tfrac\pi2$, so $I = \tfrac\pi4$. A stray factor of $x$ on $[0, \pi]$ becomes $\pi - x$; adding again gives $\int_0^\pi x\,g(\sin x)dx = \tfrac\pi2\int_0^\pi g(\sin x)dx$ — how $\int_0^\pi\tfrac{x\sin x}{1 + \cos^2x}dx = \tfrac{\pi^2}{4}$ falls out.

$\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f(2a - x) = f(x)$, and $= 0$ if $f(2a - x) = -f(x)$. Companion (sometimes "Jack's rule"): $\int_{-a}^a f = 2\int_0^a f$ for even $f$, $0$ for odd $f$. Example: $\int_0^\pi\sin^3x\,dx = 2\int_0^{\pi/2}\sin^3x\,dx = 2\cdot\tfrac23 = \tfrac43$; and $\int_0^{2\pi}\cos^3x\,dx = 0$ by the odd-symmetry about $\pi$... check: $\cos^3(2\pi - x) = \cos^3 x$ so use the period instead — $\int_0^{2\pi}\cos^3 = 0$ because $\cos^3$ has zero mean over a full period.

For any integrable $f$ and constants $a_i > 0$, $b_i$: $\displaystyle\int_{-\infty}^\infty f\!\left(x - \sum_i\frac{a_i}{x - b_i}\right)dx = \int_{-\infty}^\infty f(x)\,dx$. The simplest case (Cauchy–Schlömilch) is $\int f(x - \tfrac1x)dx = \int f(x)dx$: the map $u = x - \tfrac1x$ covers every real value once on each half-line, and the two pieces of $du$ add up to $dx$. So $\int_{-\infty}^\infty e^{-(x - 1/x)^2}dx = \int e^{-x^2}dx = \sqrt\pi$ — a hideous integrand, evaluated by recognizing the pattern.

On $(0, \infty)$ the substitution $x \to \tfrac1x$ maps the interval to itself. Two uses: it can show an integral equals its own negative (so it is $0$: $\int_0^\infty\tfrac{\ln x}{1 + x^2}dx = 0$), or produce a second form to add: $\int_0^\infty\tfrac{dx}{1 + x^4}$ becomes $\int_0^\infty\tfrac{x^2dx}{1 + x^4}$, and adding, dividing by $x^2$, and substituting $u = x - \tfrac1x$ yields $\tfrac{\pi\sqrt2}{4}$. When the interval is $(0, \infty)$, try this before anything else.

$\int_0^\infty\tfrac{\sin^2 x}{x^2}dx = \int_0^\infty\tfrac{1 - \cos x}{x^2}dx = \tfrac\pi2$ (one integration by parts each lands on $\int_0^\infty\tfrac{\sin x}{x}dx$). Substituting $x = \tan\theta$ turns $\int_0^\infty\tfrac{\ln(1+x^2)}{1+x^2}dx$ into $-2\int_0^{\pi/2}\ln\cos\theta\,d\theta = \pi\ln 2$; and Feynman with the parameter $\arctan(ax)$ gives $\int_0^\infty\tfrac{\arctan x}{x(1+x^2)}dx = \tfrac\pi2\ln 2$.

$\int_0^\infty\sin(x^2)\,dx = \tfrac12\sqrt{\tfrac\pi2}$ (the Mellin integral $\int_0^\infty t^{s-1}\sin t\,dt = \Gamma(s)\sin\tfrac{\pi s}{2}$ at $s = \tfrac12$). $\Gamma(s)\Gamma(1-s) = \tfrac{\pi}{\sin\pi s}$, which evaluates $\int_0^\infty\tfrac{dx}{1+x^n} = \tfrac{\pi/n}{\sin(\pi/n)}$. Wallis: $\prod_{n\ge1}\tfrac{4n^2}{4n^2-1} = \tfrac\pi2$, from the reduction formula for $\int_0^{\pi/2}\sin^nx\,dx$ squeezed between consecutive terms. And $\int_0^1\tfrac{\ln x}{1-x}dx = -\tfrac{\pi^2}{6}$ by expanding $\tfrac{1}{1-x}$ and integrating termwise.

$\displaystyle\int_0^\infty\frac{f(ax) - f(bx)}{x}dx = \big(f(0) - f(\infty)\big)\ln\frac ba$. With $f = e^{-t}$: $\int_0^\infty\tfrac{e^{-x} - e^{-3x}}{x}dx = \ln 3$. Proof idea: write $\tfrac{f(ax) - f(bx)}{x} = -\int_a^b f'(tx)\,dt$ and swap the order of integration — Feynman's idea once more.

$\Gamma(n+1) = \int_0^\infty x^ne^{-x}dx = n!$ (integrate by parts $n$ times), and $B(m,n) = \int_0^1x^{m-1}(1-x)^{n-1}dx = \tfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. So $\int_0^1x^3(1-x)^2dx = \tfrac{3!\,2!}{6!} = \tfrac1{60}$ instantly, and $\int_0^\infty x^4e^{-x}dx = 24$. Also $\Gamma(\tfrac12) = \sqrt\pi$, which is the Gaussian integral wearing a different hat.

$t = \tan\tfrac x2$ turns $\sin x = \tfrac{2t}{1+t^2}$, $\cos x = \tfrac{1-t^2}{1+t^2}$, $dx = \tfrac{2\,dt}{1+t^2}$: ANY rational function of $\sin$ and $\cos$ becomes a rational function of $t$, solvable by partial fractions. Example: $\int_0^\pi\tfrac{dx}{2 + \cos x} = \int_0^\infty\tfrac{2\,dt}{3 + t^2} = \tfrac{\pi}{\sqrt3}$.

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.