Monday, May 15, 2017

Unizor - Definite Integrals - Improper Integrals

Notes to a video lecture on

Improper Definite Integrals

Recall the definition of the definite integral:
abf(x) dx =
Σi∈[1,N] f(xiΔxi
where Δxi=xi−xi−1 represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δxi is shrinking to zero by length as N→∞.

Before we only defined and calculated these integrals for cases where the segment of definition [a,b] was finite and the integrated function f(x) was, at least, continuous on this segment and, as a consequence, was finite as well.

Consider now cases of integration over unbounded (infinite) domains and/or functions that go to infinity around some point(s) within their domains. Let's start with a statement that we never defined these types of integrals. Riemann sums are not applicable in these cases.

For example, you want to determine an area under the curve y=1/x² from left boundary x=1 to positive infinity.
This function diminishes to zero as x→∞, so, intuitively, the area under this infinite curve might or might not be finite, depending on how fast the function value goes to zero as its argument goes to infinity.

Our first order is to define definite integrals of this kind.

Definition for infinite intervals of integration

We will define an improper integral of each kind as a limit of the corresponding proper integral if and only if this limit exists.

1. af(x) dx =
limb→∞ abf(x) dx

provided this limit exists.

2. b−∞ f(x) dx =
lima→−∞ abf(x) dx

provided this limit exists.

When both margins are infinite, we can define the integral as a sum of two integrals, each with only one margin being infinite, provided both exists in a sense of corresponding limits as defined above.
3. −∞ f(x) dx =
0−∞ f(x) 
dx + 0 f(x) dx

Definition for functions going to infinity

Assume, we are integrating a function that asymptotically goes to infinity around one point within or on the border of a segment of integration.
For example,
01ln(x) dx
As we know, logarithm goes to negative infinity as we approach argument 0, which is a left boundary of integration segment.
To define this integral, we will cut off this point out of integration by stepping side-wise and take a limit of the result as the point of cut-off is getting closer and closer to a point where our function is not defined. If this special point is on the border of a segment of integration, we will have to take only one such limit. If it's in the middle, we will have to split the segment in two parts and integrate each one separately using this technique.

Assume, function f(x) is defined and continuous on interval (a,b] that is open on the left because f(x)→∞ as x→a.
Then we define
4. abf(x) dx =
limd→0 ba+d f(x) dx

For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Analogously, assume, function f(x) is defined and continuous on interval [a,b) that is open on the right because f(x)→∞ as x→b.
Then we define
5. abf(x) dx =
limd→0 ab−df(x) dx

For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Finally, assume, function f(x) is defined and continuous on intervals [a,b) and (b,c] - everywhere at segment [a,c] except point x=b because f(x)→∞ as x→b.
Then we define
6. acf(x) dx =
dx + bcf(x) dx

provided both integrals on the right exist in a sense of limits defined above.


Example 1

11/x² dx =
limb→∞ 1b1/x² dx

The indefinite integral (antiderivative) of f(x)=1/x² is function g(x)=−1/x. Therefore, the definite integral on the right in the above equality can be evaluated by Newton-Leibniz formula as
1b1/x² dx =
= (−1/b) − (−1/1) = 1 − 1/b

Now we can take a limit of this expression as b→∞:
limb→∞ (1 −1/b) = 1
Therefore, according to the definition of this improper integral,
11/x² dx = 1

Example 2

01ln(x) dx =
limd→0 d1ln(x) dx

Indefinite integral (antiderivative) of f(x)=ln(x) is g(x)=x·ln(x)−x (refer to a lecture about indefinite integrals or just differentiate this expression to check that its derivative is indeed equal to ln(x)).
Using Newton-Leibniz formula, we can evaluate the integral on the right:
d1ln(x) dx =
= (1·ln(1)−1) − (d·ln(d)−d) =
= −1 − d·ln(d) + d

Going to a limit as d→0, we notice that
limd→0 d·ln(d) = 0 (see a note with the proof below) and, therefore, our integral, according to the definition, is
01ln(x) dx = −1
NOTE: Proof of the limit:
limd→0 d·ln(d) =
limd→0 ln(d)/(1/d)

Use L'Hopital's rule to replace the ratio of functions with ratio of their derivatives.
limd→0 ln(d)/(1/d) =
limd→0 (1/d)/(−1/d²) =
limd→0 (−d) = 0

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