Function integral on ℝⁿ

Set

context	$p\in \mathbb N$
definiendum	$I^p: (\mathbb R^p\to\overline{\mathbb R})\to\overline{\mathbb R}$
definiendum	$I^p(f):=\int_{\mathbb R^p}\ f\ \mathrm d\lambda^p$

Discussion

Because the integral above coincides with the Lebesgue–Stieltjes integral for the monotone function $F(x):=x$ , we'll also denote $I^p(f)$ by $\int_{\mathbb R^p}\ f(x)\ \mathrm dx^p$ with the argument $x\in \mathbb R^p$ of $f$ becoming a dummy index.

Theorems

For $f:X\to \mathbb R$ …differentiable and $f'$ …bounded, we have

$\int_a^b\dfrac{{\mathrm d}f}{{\mathrm d}x}{\mathrm d}x = f(b)-f(a)$
${\mathrm d}\left(\int_{v(y)}^{w(y)}\,f(x)\,{\mathrm d}x\right) = f(v(y))\,{\mathrm d}v(y)-f(w(y))\,{\mathrm d}w(y)$

For $f$ convex and

$\langle f\rangle_{[a,b]}:=\dfrac{1}{b - a}\int_a^b f(x)\,{\mathrm d}x$

$\dfrac{f(a) + f(b)}{2} \ge \langle f\rangle_{[a,b]} \ge f \left(\dfrac{a+b}{2}\right)$

See references.

Kernel of he integral

A linear combination of functions that are zero under an integral are again zero.

Special case

$\int_{-a}^a E(x) \left( \dfrac{1}{2} + \sum_{k=0}^\infty c_k U_k(x)^{2k+1} \right) \,{\mathrm d}x = \int_0^a E(x) \,{\mathrm d}x$

e.g. all $U_k$ the same and $c_k$ so that you get $\frac{1}{1\pm e^{y}}$ :

$\int_{-a}^a E(x) \dfrac{1}{1\pm {\mathrm e}^{U(x)}}\,{\mathrm d}x = \int_0^a E(x) \,{\mathrm d}x$

$\int_{-a}^a f(x^2) \dfrac{1}{1 + {\mathrm e}^{x^2\sin(x)}}\,{\mathrm d}x = \int_0^a f(x^2) \,{\mathrm d}x$