# Difference between revisions of "Delta integral"

There are a few equivalent definitions of $\Delta$-integration.

## Cauchy $\Delta$-integral

Let $\mathbb{T}$ be a time scale. We say that $f$ is regulated if its right-sided limits exist (i.e. are finite) at all right-dense points of $\mathbb{T}$ and its left-sided limits exist (i.e. are finite) at all left-dense points of $\mathbb{T}$. We say that $f$ is pre-differentiable with region of differentiation $D$ if $D \subset \mathbb{T}^{\kappa}$, $\mathbb{T}^{\kappa} \setminus D$ is countable with no right-scattered elements of $\mathbb{T}$, and $f$ is $\Delta$-differentiable at each $t \in D$. Now suppose that $f$ is regulated. It is known that there exists a function $F$ which is pre-differentiable with region of differentiation $D$ such that $F^{\Delta}(t)=f(t)$. We define the indefinite integral of a regulated function $f$ by $$\displaystyle\int f(t) \Delta t = F(t)+C$$ for an arbitrary constant $C$.

Now we define the definite integral, i.e. the Cauchy integral, by the formula $$\displaystyle\int_s^t f(\tau) \Delta \tau = F(t)-F(s)$$ for all $s,t \in \mathbb{T}$.

A function $F \colon \mathbb{T}\rightarrow \mathbb{R}$ is called an antiderivative of $f \colon \mathbb{T}\rightarrow \mathbb{R}$ if $F^{\Delta}(t)=f(t)$ for all $t \in \mathbb{T}^{\kappa}$. It is known that all rd-continuous functions possess an antiderivative, in particular if $t_0 \in \mathbb{T}$ then $F$ defined by $$F(t) = \displaystyle\int_{t_0}^t f(\tau) \Delta \tau$$ is an antiderivative of $f$.

## Related definitions

If $a \in \mathbb{T}$, $\sup \mathbb{T}=\infty$, and $f$ is rd-continuous on $[a, \infty) \cap \mathbb{T}$ then we define the improper integral by $$\displaystyle\int_a^{\infty} f(t) \Delta t = \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_a^b f(t) \Delta t$$

## Properties of $\Delta$-integrals

Theorem: The following formula holds: $$\int_t^{\sigma(t)} f(\tau) \Delta \tau = \mu(t)f(t)$$

Proof:

Theorem: The following formula holds: $$\int_a^b [f(t)+g(t)]\Delta t = \int_a^b f(t) \Delta t + \int_a^b g(t) \Delta t$$

Proof:

Theorem: If $\alpha$ is constant with respect to $t$, then $$\int_a^b (\alpha f)(t) \Delta t=\alpha \int_a^b f(t) \Delta t.$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t) \Delta t = -\int_b^a f(t) \Delta t$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t) \Delta t = \int_a^c f(t) \Delta t + \int_c^b f(t) \Delta t.$$

Proof:

Theorem (Integration by Parts,I): The following formula holds: $$\int_a^b f(\sigma(t))g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t)g(t) \Delta t.$$

Proof:

Theorem (Integration by Parts,II): The following formula holds: $$\int_a^b f(t) g^{\Delta}(t) \Delta t = (fg)(b) - (fg)(a) - \int_a^b f^{\Delta}(t) g(\sigma(t)) \Delta t.$$

Proof:

Theorem: The following formula holds: $$\int_a^a f(t) \Delta t = 0.$$

Proof:

Theorem: If $|f(t)| \leq g(t)$ on $[a,b)$ then $$\left| \int_a^b f(t) \Delta t \right| \leq \int_a^b g(t) \Delta t$$

Proof:

Theorem: If $f(t) \geq 0$ for all $a \leq t < b$ then $$\displaystyle\int_a^b f(t) \Delta t \geq 0.$$

Proof:

Theorem (Fundamental theorem of calculus,I): The following formula holds: $$\int_a^b f^{\Delta}(t) \Delta t = f(b)-f(a).$$

Proof:

Theorem (Fundamental theorem of calculus,II): The following formula holds: $$\left( \int_{t_0}^x f(\tau) \Delta \tau) \right)^{\Delta} = f(x).$$

Proof: