Thursday, May 25, 2017

Unizor - Definite Integrals - Improper Integrals Examples 2

Notes to a video lecture on http://www.unizor.com

Improper Definite Integrals

Example 2.1

012x/(1−x²) dx =
limb→1 0b2x/(1−x²) dx =
limb→1 0b1/(1−x²) d(x²)

Substitute t=x², including the limits of integration for t, which, considering x∈[0,b], would be t∈[0,b²].
The resulting integral would be
limb→1 01/(1−t) dt

Indefinite integral of function f(t)=1/(1−t) is F(t)=−ln(1−t).
Indeed, let's take a derivative of F(t)=−ln(1−t):
Dx F(t) = −(−1/(1−t)) = 1/(1−t)

Using Newton-Leibniz formula,
01/(1−t) dx =
= F(b²) − F(0) =
= − ln(1−b²) + ln(1−0)

As b→1ln(1−b²) is decreasing to negative infinity. So, this expression is increasing to positive infinity and we conclude that the original integral diverges.

Answer: This integral diverges, it has no real value.

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Example 2.2

1sin(1/x)/x² dx =
limb→∞ 1bsin(1/x)/x² dx =
=
limb→∞ −1bsin(1/x)d(1/x)=
limb→∞ −11/bsin(t)dt =
limb→∞ 11/b d(cos(t)) =
limb→∞
[cos(1/b) − cos(1)]
As b→∞, this expression converges to 1 − cos(1)

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Example 2.3

Analyze the convergence of the following integral, depending on the value of parameter a.
1xa dx

The indefinite integral of the function f(x)=xa for any a≠−1 is F(x)=xa+1/(a+1)
The case of a=−1 should be considered separately (see below).
Convergence of our integral for all cases when a≠−1 depends on convergence of the following limit:
limb→∞[F(b)−F(1)] =
= [1/(a+1)limb→∞[ba+1−1]
Since infinitely increasing variable b in any positive power produces infinitely increasing variable, values of a that are greater than −1 should be excluded.
Infinitely increasing variable b, raised to a negative power will converge to zero.
Therefore, for all values of a that are less then −1 the limit will be −1/(a+1).
Consider now a case of a=−1.
Indefinite integral of function f(x)=x−1=1/x is F(x)=ln(x)
Since this function is infinitely growing as x→∞, the original integral does not converge to any real number.

This improper integral converges only for those values of parameter a that are less than −1, in which case the integral equals to
1xa dx = −1/(a+1)
For example, for a=−2:
1x−2 dx = 11/x² dx = 1

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Example 2.4

Analyze the convergence of the following integral, depending on the value of parameter a.
01xa dx

First of all, for all positive values of parameter a this is a proper integral and its value is
01xa dx = 1/(a+1)
For a=0 our function f(x)=xa is constant and equals to 1 everywhere except on the left margin x=0, where it is undefined.
So, we really have to calculate
limb→0b1x0 dx =
limb→0(1−b) = 1

(in which case the answer 1/(a+1) still valid).

Assume now that parameter a is negative. Obviously, in this case the function xa grows to infinity as x→0, which makes our integral improper.
The indefinite integral of the function f(x)=xa for any a≠−1 is F(x)=xa+1/(a+1)
The case of a=−1 should be considered separately (see below).
Convergence of our integral for all cases when a≠−1 depends on convergence of the following limit:
limb→0[F(1)−F(b)] =
= [1/(a+1)limb→0[1−ba+1]
Since infinitesimal variable b in any negative power produces infinitely growing variable, values of a that are less than −1 should be excluded.
Infinitesimal variable b, raised to a positive power will converge to zero.
Therefore, for all values of a that are greater then −1 the limit will be 1/(a+1).
Consider now a case of a=−1.
Indefinite integral of function f(x)=x−1=1/x is F(x)=ln(x)
Since this function is infinitely decreasing to negative infinity as x→0, the original integral does not converge to any real number.