Unit 3 | unit 3 improper integrals

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TRF

Unit - 3

Improper integrals

Q1) Define definite integrals.

A1)

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

Q2) Evaluate.

A2)

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

Q3) Evaluate

A3)

Here is an increasing function on [1, 2]

So that,

…. (1)

We know that-

And

Then equation (1) becomes-

Q4) Evaluate-

A4)

Q5) Explain improper integrals over finite intervals.

A5)

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-

Q6) What do you understand by convergence at left end?

A6)

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.

Q7) What do you understand by convergent at right-end?

A7)

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.

Q8) What is the comparison test in limit form?

A8)

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’.

Q9) Test the convergence of

A9)

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.

Q10) Show that is divergent.

A10)

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.

Q11) What is the convergence of beta function?

A11)

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.

Q12) Show that , p>0 converges absolutely for p<1.

A12)

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 .

Q13) What is the convergence at infinity?

A13)

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.

Q14) Show that is divergent.

A14)

For X > 0, we have

Here

So that is divergent.

Q15) What do you understand by the convergence of gamma function?

A15)

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 .

Q16) Define sequence and series of a function.

A16)

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.

Q17) Explain point-wise convergence.

A17)

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 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.

Q18) Prove that A sequence of differentiable functions { } with limit 0 for which {} diverges

A18)

Let

Then

But

And so

Q19) What is uniform convergence?

A19)

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.

Q20) 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.

A20)

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