Improper Integrals
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Improper Integrals Definition
let be integrable on any interval
If exists and it is a finite number, then , and is said to be convergent.
Example1:
Does exist?
Solution: so the integral is divergent
Example 2:
Check convergence of integration from 2008 to infinity 1/2dt
lim integration from 2008 to x 1/t2dt=lim (-1/x+1/2008)=1/2008
x->infinity x->infinity
so it's convergent
Recall if f,g integrable on [a,b] and f<or=g, integration from a to b f(x)dx<=integration from a to b g(x)dx
Lamprison Theorem:
If f(x)>=g(x)>=0 for any x
and f,g integrable on any interval, then
(1) If integration from a to infinity f(x)dx convergent => integration from a to infinity g(x)dx is convergent
(2) If integration from a to infinity g(x)dx is divergent => integration from a to infinity f(x)dx is divergent
Q: show that integration from 1 to infinity e^-(x2)dx is convergent for x>1
Solution:
since e^(-x2)<=e-x for any x
because integration from 0 to infinity e-xdx=lim (e-x)/(x-1)=0
x->infinity
so integration from 0 to infinity e-xdx is convergent
then integration from 1 to infinity e-x is convergent
2nd kind
Assume interval is finite but f has some discontinuity in the interval, Determine whether integration from a to b f(x)dx is convergent or not?
Answer:
If limit integration from a to (b-segma) f(x)*dx exists and is finite
segma->o, and segma>0
I= integration from 0 to 1 1/(x-1)*dx
x=1 => discontinuity for f(x)=1/(x-1)
I segma= integration from 0 to (1-segma) 1/(x-1)*dx,
segma>0
I segma=ln|x-1| from 0 to 1-segma=ln|1-segma-1|-ln|0-1|
=ln segma
as segma->0, lim I segma=lim (ln segma)=-infinity
segma-->0 segma-->0
Conclusion:
Integration from 0 to 1, 1/(x-1)dx is divergent
integration from a to b f(x) is convergent if lim integration from (a+segma) to b f(x)*dx is a finite number
What about integration from a to b f(x)dx if f is discontinuous in C. (a<c<b) either lim f(x) is not finite, or lim f(x) is not finite or both.
In this case, write integration from a to b f(x)dx=integration from a to c f(x)dx+from c to b f(x)dx
Another example:
Question: For what values of p, the integral of integration from 1 to infinity 1/xp dx is convergent
Answer:
Translate convergent:
lim(∫ 1 to A)1/x^p*dx= lim x^(1-p)/(1-p) from 1 to A
A->infinity A->infinity
= lim [A^(1-p)/(1-p)-1/(1-p)}
A->infinity
=1/1-p+1/1-p lim A^(1-p)
A->infinity
If 1-p>0, then the limit is infinite
if 1-p<0, then the limit is 0