Definite integral

Definitions

For a fixed $f : R \to R$ , the definite integral function is $R^{2} \to R$ . In general, it is defined for any $f$ over a measurable space.

Simply see the definition of min-cover integrals defined elsewhere: those form superior notions of integration than what follows.

Integral as least Upper sum, greatest lower sum

Aka Reimann Integral.

Partitions of [a,b]

Any such partition $P$ is specified by a set of points $a = x_{0} \leq x_{1} . . x_{n} = b; Δ x_{i} = x_{i} - x_{i - 1}$ .

Refinement

$P^{'}$ is a refinement of $P$ if $P \subseteq P^{'}$ . Common refinement of $P_{1}, P_{2} : P_{1} \cup P_{2}$ .

Upper, lower sums

Take bounded function \ $f : [a, b] \to R; M_{i} = sup {f (x) | x \in [x_{i - 1}, x_{i}]}; m_{i} = inf {. .}$ ; visualize; upper, lower sums $U (P, f) = \sum_{i = 1}^{n} M_{i} Δ x_{i}; L (P, f) = \sum_{i = 1}^{n} m_{i} Δ x_{i}; \bar{\int_{a}^{b}} f (x) d x = inf_{P} {U (P, f)}, \underline{\int_{a}^{b}} f (x) d x = sup_{P} {L (P, f)} $: b o t h e x i s t a s b o t h $ L $a n d $ U$ sums are bounded and real.

Integrability

If $L = U$ , $f \in R$ , that is: $f$ is Reimann integrable.

Limitations

Not Reimann integrable: $I_{Q} (x)$ (Indicator function for rational numbers): $U = 1$ , while $L = 0$ as every partition will contain a rational and an irrational number. However, this is box integrable: the corresponding box Integral is 0.

It is but a special case of superior notions such as the Stieltjes integral and the min-cover integral.

Min-cover integral of measurable functions

Aka Lebesgue integral, box integral. Take a real valued function $f$ over measure space $(X, S, m)$ . Suppose $f : X \to R$ is a measurable function from $(X, S, m) \to (R, S_{r}, m_{r})$ , where $S_{r}$ is the sigma algebra of open sets in $R$ , and $m_{r}$ is the common box measure on R.

For $f (x) \geq 0 \forall x$ , for any $E \in S$ , $\int_{E} f (x) d m$ is the min-cover measure of portion of the space $E_{f}$ in $(X \times R)$ bounded by $E$ and $f (x)$ . This measure was described for the general case in the product measure portion of the algebra survey. There is the additional restriction that the each $B_{i} = (B_{i X} \times B_{i R}) \in (S \times S_{r})$ used to cover $E_{i} \subseteq E_{f}$ is such that for any $(x, f (x)) \in E_{i}$ , its measure along $R$ , $m_{r} (B_{i R}) \geq m_{r} (f (x))$ : that is, they can be visualized as vertical boxes - with $R$ being the vertical axis.

This can be extended to any function $f$ . Let $f^{+} = 2^{- 1} (| f (x) | + f (x))$ and $f^{-} = 2^{- 1} (| f (x) | - f (x))$ be the positive and negative parts of $f$ , so that $f = f^{+} - f^{-}$ . Then, $\int_{E} f (x) d m = \int_{E} f^{+} (x) d m - \int_{E} f^{-} (x) d m$ .

Importance

It is superior to Reimann integral. The indicator function over rationals $I_{Q} (x)$ is not reimann integrable, but it is box integrable as the measure $m (Q) = 0$ .

Min-cover integral wrt to non decreasing g

Aka Lebesgue-Stieltjes integral or Radon integral.

This is defined as the min-cover integral obtained using a measure $m_{g}$ which corresponds to $g$ .

Upper and lower sums view

Aka Stieltjes integral. Take $g : [a, b] \to R$ nondecreasing. $Δ g_{i} = g (x_{i}) - g (x_{i - 1})$ . $f$ bounded. Define $U (P, f, g) = \sum_{i = 1}^{n} M_{i} Δ g_{i}$ , L(P, f, g), $\bar{\int_{a}^{b}} f (x) d g, \int_{_}$ .

If $L = U$ , $f \in R^{'} (g)$ : Stieltjes integrable.

For $\int$ to exist, $f$ and $g$ must not share any points of discontinuity: See how g is used to deal with discontinuity in $f$ .

Advantages

Suppose $f$ is $[x \in Q]$ in a certain interval X, $f \in R (g)$ where g(x) = constant over X.

If $f$ bounded on $[a, b]$ , has finite discontinuities, g conts at these points, then $f \in R (g)$ : take partition with $δ g_{i}$ small in these pts, where $| m_{i} - M_{i} |$ maybe big but bounded; thence show $U - L < ϵ$ .

Series as Stieltjes integral

Derive g from unit step functions: $g (x) = \sum_{n = 1}^{\infty} c_{n} I (x - s_{n})$ , with $c_{n} \geq 0, \sum c_{n}$ convergent. Then $\int_{a}^{b} f (x) d g = \sum c_{n} f (s_{n})$ : From $\int f d (g_{1} + g_{2}) = \int f d (g_{1}) + \int f d (g_{2})$ . $\int_{a}^{b} 1 d g = \sum c_{n}$ .

Importance

For a fixed $f, m$ , the integral can be viewed as a signed measure over $X$ . This measure is significant as it captures and extends the notion of area/ volume etc.; and as described in another chapter, it happens to be closely related to the gradient of the function.

Basic properties

Integral sandwiched between L and U

If $P \subseteq P^{'}$ , $L (P, f, g) \leq L (P^{'}, f, g), U (P^{'}, f, g) \leq U (P, f, g)$ : see geometrically. So, $\bar{\int_{a}^{b}} f (x) d g \leq \int_{_}$ : take common refinement of of $P_{1}, P_{2} : P^{'}$ , see $L (P_{1}, f, g) \leq L (P^{'}, f, g) \leq U (P^{'}, f, g) \leq U (P_{2}, f, g)$ .

$f \in R (g)$ $\equiv$ $\forall ϵ : \exists P : U (P, f, g) - L (P, f, g) < ϵ$ . $\leftarrow$ Pf: $L \leq \int_{_} \leq \int^{¯} \leq U$ . $\to$ Pf: As $\int_{_} = \int^{¯}$ , take $L (P_{1}), U (P_{2})$ close to these, take common refinement.

Closeness to any sum between L and U

So, if $s_{i}, t_{i} \in [x_{i - 1}, x_{i}], \sum | f (s_{i}) - f (t_{i}) | Δ g_{i} \leq ϵ$ . Also, $| \sum f (t_{i}) Δ g_{i} - \int_{a}^{b} f d g | < ϵ$ : both are bounded between L and U.

Continuity implies integrability

If $f$ continuous on $[a, b]$ , $f \in R (g)$ on $[a, b]$ .

Proof

As $[a, b]$ compact, $f$ uniformly cont; so $\forall η \exists d : | x - t | \leq d ⟹ | f (x) - f (t) | < η$ ; take $η : (g (b) - g (a)) η \leq ϵ$ ; take any partition P with $Δ x_{i} \leq d$ ; so $U (P, f, g) - L (P, f, g) = \sum (M_{i} - m_{i}) Δ g_{i} \leq ϵ$ .

Relationship with Uniform convergence

Take $f_{n}$ on $[a, b]$ , $f_{n} \to f$ uniformly.

If $f_{n} \in R, f \in R$ : $l t_{n \to \infty} \int f_{n} d x = \int f d x$ as $\int (f_{n} - ϵ_{n}) d x \leq \int_{_} f d x \leq \int^{¯} f d x \leq \int (f_{n} + ϵ_{n}) d x$ .

So, if $\sum f_{n} \to g$ uniformly, $\sum \int f_{n} d x = \int g d x$ .

Continuity of semi-definite integral

If $f \in R$ on $[a, b]$ : $F (x) = \int_{a}^{x} f (t) d t$ is continuous: $f$ is bounded; so $F (y) - F (x) \leq M (y - x) < ϵ$ as $y \to x$ .

Connection to derivative

Differential of integral over f

If $f$ is continuous at $c$ : : $F (x) = \int_{a}^{x} f (t) d t$ is differentiable at $c$ , $F^{'} (c) = f (c)$ .

Proof

Continuity of f states that $t - c < d ⟹ | f (t) - f (c) | < ϵ$ . So $| \frac{F (t) - F (c)}{t - c} - f (c) | = | \frac{1}{t - c} | | \int_{c}^{t} f (u) d u - f (c) (t - c) | = | \frac{1}{t - c} | | \int_{c}^{t} (f (u) - f (c)) d u | \leq ϵ$ .

Definite integral = difference in antiderivative

Anti-derivative

If $f \in R$ , and if $\exists F$ differentiable on $[a, b]$ with $F^{'} = f$ , then $F$ is called the antiderivative or the indefinite integral of $f$ .

For any $f$ , if there is an antiderivative $F$ , ${F (x) + k | k \in R}$ is also an antiderivative. (So, there are either 0 or $\infty$ antiderivatives.)

Fundamental theorem of calculus

$\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

Proof

Pick a partition $P = (x_{i})$ with $U - L < ϵ$ . Apply Mean Value Thm to get: $F (x_{i}) - F (x_{i - 1}) = f (t_{i}) Δ x_{i}$ ; add all such terms to get $F (b) - F (a) = \sum f (t_{i}) Δ x_{i} = \int_{a}^{b} f (x) d x + ϵ$ by a property seen under Stieltjes Integral.

Inter-measure Derivative

Aka Radon-Nikodym derivative. The min-cover integral may be used to define a derivative which connects two measures on the same measurable space: this is described in the chapter on measures in the Algebra survey.

Integration: evaluating definite integrals

Methods relevant (eg: transformation to polar coordinates) only for evaluating definite integrals of functions over product spaces are described elsewhere.

Also see section on important integrals for examples of application of these techniques.

Integration by parts

As definite integral equals the difference in the anti-derivative, $\int_{a}^{b} f (x) H (x) d x + \int_{a}^{b} F (x) h (x) d x = \int_{a}^{b} f (x) h (x) d x = F (b) H (b) - F (a) H (a)$ .

Thus, $\int_{[a, b]} u dv = uv]{[a, b]} - \int{[a, b]} v du$ is a useful rule when RHS is easier to integrate.

Sum, product, decomposition rules

Due to definite integral being equal to difference in the antiderivative: $\int_{a}^{b} f (x) + h (x) d g = \int_{a}^{b} f d g + \int_{a}^{b} h d g$ . $\int_{a}^{b} f d g = \int_{a}^{c} f d g + \int_{c}^{b} f d g$ .

$\int f d (g_{1} + g_{2}) = \int f d (g_{1}) + \int f d (g_{2})$ .

Transform the integral

Suppose you want to evaluate the definite integral I. Often, evaluating kI or $I^{2}$ (as in the case of the Gaussian integral) is simpler. Thence one can deduce the value of I.

Change of variables

Function composition: integrability

If $f \in R (g), f (x) \in [m, M]$ , h continuous over [m, M], then $j (x) = h (f (x)) \in R (g)$ .

\pf{Take $\forall η \exists P : U (P, f, g) - L (P, f, g) = \sum (M_{i} - m_{i}) Δ g_{i} = δ (g (b) - g (a)) \leq η$ . (Supposing $M_{i} - m_{i} \leq Δ$ .)

$h$ uniformly continuous over $[m, M]$ , so $(h (M_{i}) - h (m_{i})) \leq ϵ$ for some $ϵ$ .

Hence, $\forall \eta’ \exists P: U(P, j, g) - L(P, j, g) = \sum (M_{i}^{} - m_{i}^{}) \gD g_{i} = \eps (g(b) - g(a)) \leq \frac{\eta \eps}{\del} = \eta’$.}

So, if $f, g \in R (g) : f g = 4^{- 1} ((f + g)^{2} - (f - g)^{2}), | f (x) | \in R (g)$ . Also, $\int f (x) d g \leq \int | f (x) | d g$ .

Integration

Take strictly increasing onto $h : [A, B] \to [a, b]$ . Suppose that $f \in R$ on $[a, b]$ .

Then $g = f \circ h \in R (h)$ on $[A, B]$ . Then $\int_{x \in [A, B]} g (x) d h = \int_{t \in [h (A), h (B)]} f (t) d h = \int_{x \in [A, B]} f (h (x)) h^{'} (x) d x$ .

\core{Take a partition $P$ with upper and lower sums $L, U$ . Use mean value theorem to rewrite in terms of $x$ .}

Utility

Change of variables is in general a extremely useful algebraic trick - so is this particular instance. Suppose $x = h (y)$ . With this, one can write: $\int_{[A, B]} f (x) d x = \int_{y \in [h^{- 1} (A), h^{- 1} (B)]} f (h (y)) \frac{d x}{d y} d y$ , as the RHS may be easier to evaluate.

Reversing limits

By convention $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ .

Over product measure spaces

Consider $f (x) \geq 0 \forall x$ . (The integral for the general case may thence be derived as earlier.)

Let $(X, S, m)$ be the product measure space of ${(X_{i}, S_{i}, m_{i}) : i \in 1, 2}$ . Let $E \in S = S_{1} \times S_{2}$ : Note that this includes only rectangles in $R^{2}$ , but not circles etc.. $E_{1} = {x_{1} \in X_{1} : \exists x = (x_{1}, x_{2}) \in E}$ . Let $E_{2} (x_{1})$ be similarly defined.

Then, from the properties of the product measure: (Fubini) $\int_{E} f (x) d m = \int_{E_{1}} g (x_{1}) d m_{1}$ , where $g (x_{1}) = \int_{x_{2} \in E_{2}} f (x_{1}, x_{2}) d (m_{2})$ .

Proof

Let $R$ be associated with the usual measure space $(R, S_{r}, m_{r})$ . (Core intuition) The min cover measure of the space in $(X \times R)$ bounded by $E$ and $f$ is $i n f_{{B_{i}}} (m \times m_{r}) (\cup_{i} B_{i})$ , where $B_{i} = (B_{i X_{1}} \times B_{i X_{2}} \times B_{i R}) \in (S \times S_{r})$ cover $E$ . Because we are dealing with the product measure, $m_{1} m_{2} m_{r} (B_{i}) = m_{1} (B_{i X_{1}}) m_{2} m_{1} (B_{i X_{2}} \times B_{i R})$ .

Thus, integration over the product measure space reduces to integrating over one variable at a time; and the order in which these integrals are taken does not matter in this case. This case is aka multiple integration.

Bounds

If $f \leq h, \int f d g \leq \int h d g$ . If $| f (x) | < M ⟹ | \int_{a}^{b} f d g | \leq M (g (b) - g (a))$ .

Use polar coordinates

Consider $\int_{a, b} f (x, y) d x d y$ , one can find an equivalent expression in polar coordinates using $x = r \sin θ, y = r \cos θ$ and $d x d y = d r (r d θ)$ . This is sometimes simpler to solve.

As in the case of the Gaussian integral, the form $\int_{a, b} f (x, y) d x d y$ may be derived from considering $I^{2}$ .

Important integrals

Normal integral

Aka Gaussian integral. $\int_{- \infty}^{\infty} e^{- x^{2}} d x = I = \sqrt{π}$ : \

Proof

$I^{2} = \int_{y = - \infty}^{y = \infty} \int_{x = - \infty}^{x = \infty} e^{- x^{2} - y^{2}} d x d y$ ; thence transform to polar coordinates and solve.

Importance

The Gaussian integral itself is important due to this nice integrability and the nice properties observed in the Gaussian distribution (very suitable for modeling). Further, it is useful because realated integrals can be used to smooth functions for various purposes (eg: optimization).

In calculating similar integrals, one often uses similar techniques and results about exponential and gamma probability densities.