| context | $p\in \mathbb N$ |
| definiendum | $I^p: (\mathbb R^p\to\overline{\mathbb R})\to\overline{\mathbb R}$ |
| definiendum | $I^p(f):=\int_{\mathbb R^p}\ f\ \mathrm d\lambda^p$ |
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.
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.
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 $\frac{1}{1\pm e^{y}}$:
$$\int_{-a}^a E(x) \dfrac{1}{1\pm {\mathrm e}^{U(x)}}\,{\mathrm d}x = \int_0^a E(x) \,{\mathrm d}x$$
$$\int_{-a}^a f(x^2) \dfrac{1}{1 + {\mathrm e}^{x^2\sin(x)}}\,{\mathrm d}x = \int_0^a f(x^2) \,{\mathrm d}x$$
Wikipedia: Hermite–Hadamard inequality