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5.4 We often integrate over an infinite range. Such integrals are called improper . They are defined a s
limits,
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−∞}^{∞}f(x)\kern 1.66702pt dx ={\mathop{ lim}}_{
a→−∞}{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{a}^{b}f(x)\kern 1.66702pt dx\quad .
Example 5.7:
Evaluate {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−∞}^{0}{e}^{x}\kern 1.66702pt dx .
Solution:
Apply the definition
\eqalignno{
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−∞}^{0}{e}^{x}\kern 1.66702pt dx ={\mathop{ lim}}_{
a→−∞}{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{a}^{0}{e}^{x}\kern 1.66702pt dx ={\mathop{ lim}}_{
a→−∞}(1 − {e}^{a}) = 1\quad . & &
}
A inite integral is called convergent, if tyhe limit does not exist the integral is called divergent.
Let us look at a physics example
Example 5.8:
Determine the escape velocity from earth.
Solution:
We need belance of energies. The initial kinetic energy must equal the work done against gravity
to get the object )of mass m
to escape from the gravity field of the earth (mass M ,
radius R )
{1\over
2}m{v}^{2} ={\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{R}^{∞}{GmM\over
{r}^{2}} dr
Evaluate the integral as above,
\eqalignno{
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{R}^{∞}{GmM\over
{r}^{2}} dr & = gmM{\mathop{lim}}_{a→∞}{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{R}^{a} {1\over {
r}^{2}}dr & &
\cr
& = gmM{\mathop{lim}}_{a→∞}\left (−{1\over
a} + {1\over
R}\right ) & &
\cr
& = {gmM\over
R} & &
}
5.4.1 Integrals that require special attentions is those where the integrand diverges. We need to take a
start-point just above and below the singularity, and take a limit. A simple and obvious example
is
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{1} {1\over
x}dx ={\mathop{ lim}}_{ϵ↓0}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{ϵ}^{1} {1\over
x}dx ={\mathop{ lim}}_{ϵ↓0}{[\mathop{ln}\nolimits (x)]}_{ϵ}^{1} ={\mathop{ lim}}_{
ϵ↓0}(−\mathop{ln}\nolimits ϵ) = ∞.
Be extrememly careful when the singularity occurs in the middle of the integration interval.
Example 5.9:
Calculate {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{1} {1\over {
x}^{2}}dx
Solution:
Split the interal into two parts,
\eqalignno{
{\mathop{lim}}_{ϵ↓0}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{−1}^{−ϵ} {1\over {
x}^{2}}dx +{\mathop{ lim}}_{δ↓0}{\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{δ}^{1} {1\over {
x}^{2}}dx & &
\cr
2 +{\mathop{ lim}}_{ϵ↓0}{1\over
ϵ} +{\mathop{ lim}}_{ϵ↓0}{1\over
δ} = ∞. & &
}
The naive answer is 2! So we note that we have to be extremely careful