What is integrable function?

Example 10.13

$\mathbf 1_\Q$ is nowhere continuous: not Riemann-integrable — but Lebesgue-trivial: $\int\mathbf 1_\Q\,\dd\lambda = \lambda(\Q) = 0$. Thomae’s function ( $\frac1q$ at rationals $\frac pq$, $0$ elsewhere) is continuous exactly at the irrationals: Riemann-integrable with integral $0$. And improper Riemann integrals are a different notion: $\int_0^\infty\frac{\sin x}x\,\dd x$ converges as a limit of $\int_0^A$ (the weekend problem computes it $= \frac\pi2$), but $\frac{\sin x}x \notin L^1(\intoo0{+\infty})$: the absolute integral diverges like the harmonic series (Exercise 10.6). Lebesgue’s theory trades conditional convergence for robust limit theorems.

Example 10.16 (The Gamma function)

For $t > 0$ let

The integral converges: near $0$, $x^{t-1}$ is integrable ($t > 0$); at infinity, $x^{t-1}\eu^{-x} \leq C\eu^{-x/2}$. Integration by parts (on $[\varepsilon, A]$, then limits via MCT) gives the functional equation $\Gamma(t + 1) = t\,\Gamma(t)$, whence $\Gamma(n+1) = n!$: the factorial interpolated. On every $\intcc ab \subseteq \intoo0{+\infty}$, $\partial_t\bigl(x^{t-1}\eu^{-x}\bigr) = \ln x\cdot x^{t-1}\eu^{-x}$ is dominated by $\abs{\ln x}(x^{a-1} + x^{b-1})\eu^{-x}$, integrable: $\Gamma$ is $\mathcal C^1$, and by induction $\mathcal C^\infty$, with $\Gamma^{(k)}(t) = \int_0^\infty(\ln x)^kx^{t-1}\eu^{-x}\dd x$. The value $\Gamma(\frac12) = \sqrt\pi$ is the Gaussian integral in disguise (Problem 10.1).