Unit - 3

Improper integrals

3.1 Improper integrals: Convergence of Beta and Gamma functions

Definite integrals-

When we apply limits in indefinite integrals are called definite integrals.

If an expression is written as , here ‘b’ is called upper limit and ‘a’ is called lower limit.

If f is an increasing or decreasing function on interval [a , b], then

Where 

Properties-

1. The definite integral applies only if a<b, but it would be appropriate to include the case a = b and a>b as well, in that case-

If a = b, then

And if a>b, then

2. Integral of a constant function-

3. Constant multiple property-

4. Interval union property-

If a < c < b, then

5. Inequality-

If c and d are constants such that for all x in [a , b], then

c(b – a)

Note- if a function f:[a , b]→R is continuous, then the function ‘f’ is always Integrable.

Example-1: Evaluate.

Sol. Here we notice that  f:x→cos x  is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

Example-2: Evaluate

Sol. Here is an increasing function on [1 , 2]

So that,

  …. (1)

We know that-

And

Then equation (1) becomes-

Note- we can find the definite integral directly as-

Example-3:  Evaluate-

Sol.         

Improper integrals

(1) Let f is function defined on [a, ∞) and it is integrable on [a , t] for all t >a, then

If exists, then we define the improper integral of f over [a, ∞) as follows-

(2) Let f is function defined on (-∞,b] and it is integrable on [t , b] for all t >b, then

If exists, then we define the improper integral of f over (-∞ , b] as follows-

(3) Let f is function defined on (-∞, ∞] and it is integrable on [a , b] for every closed and bounded interval [a , b] which is the subset of R., then

If and exist for some c belongs to R , then we define the improper integral of f over (-∞ ,∞ ) as follows-

= +

(4) Let f is function defined on (a ,∞) and exists for all t>a , then

If exists , then we define the improper integral of f over (a , ∞) as follows-

(5) Let f is function defined on (-∞ , b) and exists for all t<b  , then

If exists, then we define the improper integral of f over (-∞ , b) as follows-

Improper integrals over finite intervals-

(1) Let f is function defined on (a, b] and exists for all t ∈(a,b)  , then

If exists, then we define the improper integral of f over (a , b] as follows-

(2) Let f is function defined on [a, b) and exists for all t ∈(a,b)  , then

If exists, then we define the improper integral of f over [a , b) as follows-

(3) Let f is function defined on [a, c) and (c, b] . If and exist then we define the improper integral of f over [a, b] as follows-

We will read more about the uses of improper integrals in the next topic.

Integration of Unbounded Function with finite limits of integration-

Let a function f be defined in an interval [a, b] everywhere except possible at finite number of points.

Convergence at left –end-

Let a be the only points of infinite discontinuity of f so that according to assumption made in the last section, the integral

𝑇ℎ𝑒 𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 is defined as the

So that

If this Limit exists and is finite, the improper integral is said to converge at (a) if otherwise, it is called divergent.

Convergence at right-end

Suppose b be the only point of infinite discontinuity the improper integral is then defined by the relation

If the limit exists, the improper integral is said to be convergent at 𝑏. Otherwise

Is called divergent.

Convergence at both the end points-

If the end points a and b are the only points of infinite discontinuity of f , then for any

Point c, a < c < b,

Convergence at Interior points-

If an interior point c, a < c < b, is the only point of infinite discontinuity of f, we get

Comparison test for the convergence and divergence

If f and g are two positive functions and ‘a’ is only singular point of f and g on [a. b], such that

f(x) ≤𝑔(𝑥), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥∈[𝑎,𝑏]

1. converges if converges.
2. diverges, if converges.

Comparison test in limit form

If f and g are two positive functions [a, b] and ‘a’ is  the only singular point of f and g in [a, b], such that

𝑙𝑖𝑚𝑥→𝑎+(𝑥)𝑔(𝑥) = l

Where ‘l’ is a non – zero finite number.

Then, the two integrals and converges and diverges

Together at ‘a’.

Note- The improper integral Converges if and only  if n < 1.

Example: test the convergence of

Sol:

Here we have

Here we can see that is a bounded function.

Suppose M is its upper bound then

Also since

Is convergent as n = ½ < 1.  Therefore,

Is convergent.

Example: Show that is divergent.

Sol:

Let us suppose

Here x=1 is only singular point.

Take (𝑥) =1/x-1

Then

Thus, and are same.

Since is divergent, hence is divergent.

Convergence of beta and gamma functions

Beta function- Show that

Exists if and only if m and n both are positive.

Proof:

It is a proper integral for 𝑚≥1,≥1 ,0 and 1 are the only points of infinite discontinuity; 0 when m < 1 and 1.  When n < 1, we have

Converges at 0, when m < 1,

Suppose

Take g(x) =

Then

Since
converges iff. 1 – m < 1 or m > 0.

Thus

converges for m > 0.

Converges at x = 1,

We take

Take

Then

Also converges if and only if 1−n < 1 or n > 0.

Thus, converges if n > 0.

Hence converges if m > 0 , n > 0 .

Absolute convergence

The improper integral is said to be absolutely convergent if is convergent

Note- Every absolutely convergent integral is convergent.

Example: Show that , p>0 converges absolutely for p<1.

Sol:

Suppose

‘0’ is the only point of infinite discontinuity and f does not keeps the same sign in [0, 1].

So that

Also converges for 𝑝 < 1.

Thus

  converges if and only if p > 0.

Hence

is absolutely convergent if and only if 𝑝 < 1.

Convergent at infinity

The symbol , 𝑥≥ 𝑎 is defined as limit of when x tends to infinity, so that

If the limit exists and is finite then the improper integral (1) is said to be divergent.

Example: Show that is divergent.

Sol:

For X > 0, we have

Here

So that is divergent.

Gamma function-

The integral

Is convergent if and only if 𝑚>0.

Proof:

Suppose

If 𝑚 < 1, the ‘0’ infinite discontinuity.

So we need to examine the convergence of above improper integral at both 0 and ∞.

Convergence at 0 for 𝒎<1:

Let

Then

Since converges, if and only if m>0.

Therefore converges if and only if 𝑚>0.

Convergence at

Let

So that

Since is divergent

Thus is convergent for every m.

Hence is convergent if and only if 𝑚> 0 and is denoted by < 𝑚) .

Thus

Thus Γ(0), Γ(-1), etc. are not exists .

Key takeaways:

1. If an expression is written as , here ‘b’ is called upper limit and ‘a’ is called lower limit.
2. If a function f:[a , b]→R is continuous, then the function ‘f’ is always Integrable.
3. 𝑇ℎ𝑒 𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 is defined as the

4.     If f and g are two positive functions and ‘a’ is only singular point of f and g on [a. b], such that

f(x) ≤𝑔(𝑥),𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥∈[𝑎,𝑏]

converges if converges.

diverges, if converges.

The improper integral Converges if and only  if n < 1.

5.     The improper integral is said to be absolutely convergent if is convergent

6.     Every absolutely convergent integral is convergent

3.2 Pointwise and uniform convergence of sequence of functions

Sequence and series of a function:

Let be a real valued function defined on an interval I and for each , then < is called a sequence of real valued function on I.

We denote it by {} or <>

If > is a sequence of real valued function on an interval I, then is called the series of real valued function defined on an interval I.

This series is denoted by

That is, we shall consider sequences whose terms are functions rather than real numbers. These sequences are useful in obtaining approximations to a given function.

Point-wise convergence-

Let be a sequence of functions from a set X to R.

We say that converges to f pointwise on X if for each the sequence ( of real numbers converges to the real number f(x) in R.

Like we say a sequence ( is convergent on X. We may also define a sequence of function (is pointwise convergent on X.

This means that there exists a function such that ( is pointwise convergent to f. There is an another problem for us. We need to find the limit function f and then show that pointwise.

This means that we fix first and form the sequence of real numbers.

For any given , we have to find an such that we have |

Thus may depend not only on but also on a.

Example: A sequence of continuous functions whose limit function is discontinuous.

Let

We note that

Each is continuous on R but the limit function f is discontinuous at x = 1 and x = -1.

Example: A double sequence in which limit process cannot be interchanged:

For m =1, 2,….,

n = 1, 2,3,..., let us consider the double sequence

For every fixed n, we have

And so

On the other hand, for every fixed m, we have

And so

Hence

Example: A sequence of differentiable functions { } with limit 0 for which {} diverges

Let

Then

But

And so

Uniform convergence-

A sequence of function { is said to converge uniformly to a function f on a set E if for every there exists and integer N such that n > N implies

If each tern of the sequence < is real-valued, then the expression (1) can be written as

Definition:  A series is said to be uniformly convergent on E if the sequence of partial sums defined by converges uniformly on E.

Note- Every uniformly convergent sequence is pointwise convergent but not conversely.

Example: Consider the sequence < defined by

Then

Hence converence pointwise to 0 for all 0 < x < 1.

Let be given, then for convergence, we have

If is taken as integer greater than 1/x, then

Since depends both on in (0, 1), so does not converge uniformly on (0, 1).

Example: Consider the sequence< defined by

Then

Then < converges pointwise to 0 for all Let then for convergence we must have

If is taken as integer greater than 1/, then

Hence converges uniformly to f on [0,

Example: Consider the sequence < defined by

Then

Let be given. Then for convergence, we must have

Thus we should take to be an integer next higher to .

If we take x = 1 then m does not exists.

Thus the sequence is not uniformly convergent to f in the interval which contains 1.

Cauchy’s principal of uniform convergence

The necessary and sufficient condition for a sequence of functions  defined on A to converge uniformly on A is that for every > 0, there exists a positive integer m such that

| (x) –(x)| <   for n > k m and " x A

Key takeaways:

1. Let be a real valued function defined on an interval I and for each , then < is called a sequence of real valued function on I.
2. We say that converges to f pointwise on X if for each the sequence ( of real numbers converges to the real number f(x) in R.
3. A series is said to be uniformly convergent on E if the sequence of partial sums defined by converges uniformly on E.
4. Every uniformly convergent sequence is pointwise convergent but not conversely.

3.3 Theorems on continuity, derivability and integrability of the limit function of a sequence of functions

Uniform Convergence and Continuity

Definition- A sequence of functions on D is uniformly convergent on D if and only if, for all > 0, there exists a N ∈ N such that |fn(x) − f(x)| < for all x ∈ D and n ≥ N.

Definition of supremum norm-

Suppose f : D → R is a function with domain D. We define the sup-norm of f over D as:

a function f is bounded if and only if ||f||  < ∞, which follows immediately from the definitions. The connection between sup-norm and uniform convergence is also straight-forward.

Theorems on continuity:

Theorem-1: If f, g are two continuous functions at a point c then the functions f + g, f – g and fg are also continuous at c and g(c) is non-zero then f/g is also continuous at c.

Theorem-2: A function f is defined on an interval I, is continuous at a point c belongs to I, is continuous at point iff for every sequence converging to c, we have

Proof:

First let us suppose that the function f is continuous at a point , and { is a sequence such that
 

Since f is continuous at c, therefore for any given such that

Again since , therefore for every, a positive integer m, such that

From equation (1), put , we have

The sequence { converges to f(c).

Now suppose that  that f is not continuous at c.

We will now show that through a sequence { converging to c yet the sequence { does not converges to f(c).

Since f is not continuous at c, therefore there exists an .

Such that for every , an x such that

Also

By putting , we find that for each positive integer n,

There is an such that

Thus the sequence { does not converge to f(c).

Example: Examine the continuity at the origin

Sol:

Now,

Thus

Also

Thus the function is continuous at origin.

Theorem: If a function f is continuous on a closed interval [a, b], then it attains its bounds at least once in [a., b]

Proof:

If f is a continuous function, then it attains its bounds at every point of the interval.

Suppose f is a function which is not constant.

Since f is continuous on the interval which is closed [a, b], so that it is bounded.

Let m and M are the infimum and supremum of f.

It is to be shown that for every points p and q of [a, b] such that

f(p) = m and f(q) = M

Now consider the supremum case-

Let f does not attains the supremum M so that the function does not take the value M for any point

Which means

Now consider

Which is positive for all values of x in [a, b].

Evidently the function g is continuous and so bounded in [a, b].

Let k(>0) be its supremum

So that

Which is a contradiction that M is the supremum of f in [a, b].

Hence our superposition that f does not attains the value M leads to a contradiction and hence f attains its supremum for at least one value in [a, b].

Similarly we can show that the function also attains its infimum m.

Therefore the function attains its bounds at least once in [a, b].

Theorem: if a function f is continuous at an interior point c of [a, b] and such that f(x) has the same sign as f(c) for every

Proof:

Since the function f is continuous at an interior point c of [a, b], therefore for any such that

Or

When f(c) > 0, taking to be less than f(c) we find that

When f(c) > 0, taking to be less than -f(c) we find that

Theorem: if a function f is continuous on [a, b] and , then it assumes every value between f(a) and f(b).

Proof:

Suppose A is any number between f(a) and f(b). Now we show that there exists a number such that f(c) = A.

Let consider a function defined on [a, b] such that

Hence it it clear is continuous on [a, b].

Also

And

So that and has positive signs.

Thus the function is continuous on [a, b] and and are of opposite signs, therefore for every , such that

Theorem- Let be a sequence of continuous functions on a set E R and suppose that converges uniformly on E to a function f : E R . Then the limit function f is continuous.

Proof:

Suppose cE be an arbitrary point. If c is an isolated point of E, then f is automatically continuous at c. So suppose that c is an accumulation point of E. We shall show that f is continuous at c. Since uniformly, for every > 0 there is an integer N such that n N  implies

| (x) –f(x)| < for all x E

Since is continuous at c, there is a neighbourhood such that x (since c is limit point) implies

| (x) –(c)| <

By triangle inequality, we get

Hence

Which is the proof of continuity of f at arbitrary point .

Theorem- If the sequence of continuous function is uniformly convergent to a function f on [a, b] then f is continuous on [a, b].

Proof:

Suppose be given.

The sequence is uniformly convergent to f on [a, b], then there exists a positive integer m such that

| (x) –f(x)| <   for every ….(1)

Let be any point of [a, b].

In particular then from (1),

| () –f()| <   for every ……(2)

Now is continuous at

So there exists such that

| () –f()| < whenever |x -   …….(3)

Hence for |x - , we have

From (1), (2) and (3)

We get |(x) – f()| < whenever |x -  

Hence f is continuous at

f is continuous on [a, b].

Theorem-1:

Uniform convergence and continuity-

If be a sequence of continuous functions defined on [a, b] and f uniformly on [a, b], then

f is continuous on [a, b].

Theorem -2:

Uniform Convergence and Differentiation

Let be a sequence of functions, each differentiable on [a, b] such that converges for some point of [a, b]. If converges uniformly on [a, b] then converges uniformly on [a, b] to a function f such that

Theorem-3:

Uniform Convergence and Integration

If a sequence converges uniformly to f on [a, b] and each function is integrable on [a, b], then f is integrable on [a, b] and

Theorems on derivability-

In the function f, g are derivable at c then the functions f + g, f – g, fg and f/g where g(c) is non-negative are also derivable at c.

Theorem: if f is derivable at c and f(c) is non-zero then the function 1/f is also derivable thereat, and

Proof:

Since f is derivable at c, it is also continuous thereat.

Again since , there exists a neighbourhood N of c wherin f does not vanish.

Now

Finding limits when x tends to c, we get

Hence the limit exists and equals to .

Key takeaways:

1. Let be a sequence of continuous functions on a set E R and suppose that converges uniformly on E to a function f : E R . Then the limit function f is continuous.
2. If the sequence of continuous function is uniformly convergent to a function f on [a,b] then f is continuous on [a,b]

References:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. Advanced engineering mathematics, by HK Dass

3. Mathematical analysis by Dr. Anju panwar

4. Real analysis by Dr. Sachin Kaushal

5. Real analysis by Ajit kumar & S.Kumaresan

6. Principles of real analysis by S.C Malik

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