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

__Improper Definite Integrals__

*Example 2.1*

**∫**d_{0}^{1}2x/(1−x²)**x =**

=lim=

_{b→1}**∫**d_{0}^{b}2x/(1−x²)**x =**

=lim=

_{b→1}**∫**d_{0}^{b}1/(1−x²)**(x²)**Substitute

*, including the limits of integration for*

**t=x²***, which, considering*

**t***∈[*

**x***], would be*

**0,b***∈[*

**t***].*

**0,b²**The resulting integral would be

*lim*

_{b→1}**∫**d_{0}^{b²}1/(1−t)**t**Indefinite integral of function

**f(t)=1/(1−t)***.*

**F(t)=−ln(1−t)**Indeed, let's take a derivative of

*:*

**F(t)=−ln(1−t)***D*

_{x}**F(t) = −(−1/(1−t)) = 1/(1−t)**Using Newton-Leibniz formula,

**∫**d_{0}^{b²}1/(1−t)**x =**

= F(b²) − F(0) =

= − ln(1−b²) + ln(1−0)= F(b²) − F(0) =

= − ln(1−b²) + ln(1−0)

As

*,*

**b→1***is decreasing to negative infinity. So, this expression is increasing to positive infinity and we conclude that the original integral diverges.*

**ln(1−b²)***Answer*: This integral diverges, it has no real value.

__________

*Example 2.2*

*[*

**∫**d_{1}^{∞}sin(1/x)/x²**x =**

=lim=

_{b→∞}**∫**d_{1}^{b}sin(1/x)/x²**x =**

=lim=

_{b→∞}**−∫**d_{1}^{b}sin(1/x)**(1/x)=**

=lim=

_{b→∞}**−∫**d_{1}^{1/b}sin(t)**t =**

=lim=

_{b→∞}**∫**d_{1}^{1/b}**(cos(t)) =**

=lim=

_{b→∞}*]*

**cos(1/b) − cos(1)**As

*, this expression converges to*

**b→∞**

**1 − cos(1)***Answer*:

**1 − cos(1)**__________

*Example 2.3*

Analyze the convergence of the following integral, depending on the value of parameter

*.*

**a**

**∫**d_{1}^{∞}x^{a}**x**The indefinite integral of the function

*for any*

**f(x)=x**^{a}*is*

**a≠−1**

**F(x)=x**^{a+1}/(a+1)The case of

*should be considered separately (see below).*

**a=−1**Convergence of our integral for all cases when

*depends on convergence of the following limit:*

**a≠−1***lim*[

_{b→∞}*] =*

**F(b)−F(1)**= [

*]·*

**1/(a+1)***lim*[

_{b→∞}*]*

**b**^{a+1}−1Since infinitely increasing variable

*in any positive power produces infinitely increasing variable, values of*

**b***that are greater than*

**a***should be excluded.*

**−1**Infinitely increasing variable

*, raised to a negative power will converge to zero.*

**b**Therefore, for all values of

*that are less then*

**a***the limit will be*

**−1***.*

**−1/(a+1)**Consider now a case of

*.*

**a=−1**Indefinite integral of function

*is*

**f(x)=x**^{−1}=1/x

**F(x)=ln(x)**Since this function is infinitely growing as

*, the original integral does not converge to any real number.*

**x→∞***Answer*:

This improper integral converges only for those values of parameter

*that are less than*

**a***, in which case the integral equals to*

**−1**

**∫**d_{1}^{∞}x^{a}**x = −1/(a+1)**For example, for

*:*

**a=−2***=*

**∫**d_{1}^{∞}x^{−2}**x***= 1*

**∫**d_{1}^{∞}1/x²**x**__________

*Example 2.4*

Analyze the convergence of the following integral, depending on the value of parameter

*.*

**a**

**∫**d_{0}^{1}x^{a}**x**First of all, for all positive values of parameter

*this is a proper integral and its value is*

**a**

**∫**d_{0}^{1}x^{a}**x = 1/(a+1)**For

*our function*

**a=0***is constant and equals to*

**f(x)=x**^{a}*everywhere except on the left margin*

**1***, where it is undefined.*

**x=0**So, we really have to calculate

*lim*

lim

_{b→0}**∫**d_{b}^{1}x^{0}**x**=lim

_{b→0}**(1−b) = 1**(in which case the answer

*still valid).*

**1/(a+1)**Assume now that parameter

*is negative. Obviously, in this case the function*

**a***grows to infinity as*

**x**^{a}*, which makes our integral improper.*

**x→0**The indefinite integral of the function

*for any*

**f(x)=x**^{a}*is*

**a≠−1**

**F(x)=x**^{a+1}/(a+1)The case of

*should be considered separately (see below).*

**a=−1**Convergence of our integral for all cases when

*depends on convergence of the following limit:*

**a≠−1***lim*[

_{b→0}*] =*

**F(1)−F(b)**= [

*]·*

**1/(a+1)***lim*[

_{b→0}*]*

**1−b**^{a+1}Since infinitesimal variable

*in any negative power produces infinitely growing variable, values of*

**b***that are less than*

**a***should be excluded.*

**−1**Infinitesimal variable

*, raised to a positive power will converge to zero.*

**b**Therefore, for all values of

*that are greater then*

**a***the limit will be*

**−1***.*

**1/(a+1)**Consider now a case of

*.*

**a=−1**Indefinite integral of function

*is*

**f(x)=x**^{−1}=1/x

**F(x)=ln(x)**Since this function is infinitely decreasing to negative infinity as

*, the original integral does not converge to any real number.*

**x→0***Answer*:

This improper integral converges only for those values of parameter

*that are greater than*

**a***, in which case the integral equals to*

**−1**

**∫**d_{0}^{1}x^{a}**x = 1/(a+1)**For example, for

*:*

**a=−1/2***=*

**∫**d_{0}^{1}x^{−1/2}**x***= 2*

**∫**d_{0}^{1}1/√x**x**
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