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

__Improper Definite Integrals__

*Example 1.1*

**∫**d_{0}^{1}1/√x**x =**

=lim=

_{a→0}**∫**d_{a}^{1}1/√x**x**Indefinite integral of

**f(x)=1/√x***.*

**F(x)=2√x**Indeed, let's take a derivative of

*:*

**F(x)=2√x***D*

_{x}**F(x) = 2·(1/2)·x**

= x^{(1/2)−1}== x

^{−1/2}= 1/√xUsing Newton-Leibniz formula,

**∫**d_{a}^{1}1/√x**x = F(1) − F(a) =**

= 2√1 − 2√a= 2√1 − 2√a

As

*, this expression converges to*

**a→0**

**2***Answer*:

**2**__________

*Example 1.2*

**∫**d_{0}^{∞}e^{−x}**x =**

=lim=

_{b→∞}**∫**d_{0}^{b}e^{−x}**x**Indefinite integral of

*is*

**f(x)=e**^{−x}*.*

**F(x)=−e**^{−x}Evaluating integral:

**∫**d_{0}^{b}e^{−x}**x = F(b) − F(0) =**

= (−e= (−e

^{−b}) − (−e^{−0}) = 1 −e^{−b}As

*, this expression converges to*

**b→∞**

**1***Answer*:

**1**__________

*Example 1.3*

**∫**d_{0}^{∞}1/(1+x²)**x =**

=lim=

_{b→∞}**∫**d_{0}^{b}1/(1+x²)**x**Indefinite integral of

*is*

**f(x)=1/(1+x²)***.*

**F(x)=arctan(x)**Evaluating integral:

**∫**d_{0}^{b}1/(1+x²)**x = F(b) − F(0) =**

= arctan(b) − arctan(0) =

= arctan(b)= arctan(b) − arctan(0) =

= arctan(b)

As

*, this expression converges to*

**b→∞**

**π/2***Answer*:

**π/2**__________

*Example 1.4*

**∫**d_{1}^{∞}(1/x²)·e^{1/x}**x =**

=lim=

_{b→∞}**∫**d_{1}^{b}(1/x²)·e^{1/x}**x**Indefinite integral of

*can be found by noticing that*

**f(x)=(1/x²)·e**^{1/x}*is a derivative of*

**−1/x²***and, therefore, a derivative of*

**1/x***is*

**e**^{1/x}*.*

**e**^{1/x}·(−1/x²)Therefore, indefinite integral of

*is*

**f(x)=(1/x²)·e**^{1/x}*.*

**F(x)=−e**^{1/x}Evaluating integral:

**∫**d_{1}^{b}(1/x²)·e^{1/x}**x = F(b)−F(1) =**

= (−e= (−e

^{1/b}) − (−e^{1/1}) = e − e^{1/b}As

*, this expression converges to*

**b→∞**

**e −1***Answer*:

**e −1**
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