Can i do this to prove convergence of improper integral

In summary, the conversation discusses a method for determining whether an improper integral from 0 to 1 converges or diverges. The conditions for finding a similar function that is bigger or smaller than the given function are mentioned, but it is noted that neither condition can be applied in this case. The possibility of using the third condition is also discussed, but it is uncertain if it can be applied since the integral is not to infinity. The conversation concludes with a question about finding a suitable function to compare with the given integral near x = 0.
  • #1
Dell
590
0
given the improper integral from 0 to 1 of

[tex]\int[/tex]dx/[tex]\sqrt[3]{x}[/tex](ee-e-x)

i am asked if it comverges or diverges,

what i have learned is that if i can find :
-a similar function that is bigger than my function and that i know converges, then my function also converges
-a similar function that is smaller than my function and that i know dicerges, then my function also diverges
-a function g(x) such that
lim f(x)/g(x)=K, (K not 0, not infinity) then f(x) acts like g(x)
x->inf

my similar function g(x) is 1/3[tex]\sqrt[3]{x}[/tex] which is smaller than f(x), BUT my g(x) converges, so that doesn't prove anything according to the 1st 2 conditions,

generally can i use the 3rd condition even though my integral is from 0 to 1 and not to infinity??
in this case that too doesn't work. what can i do from here??

if not how would you solve this?
 
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  • #2
Dell said:
given the improper integral from 0 to 1 of

[tex]\int[/tex]dx/[tex]\sqrt[3]{x}[/tex](ee-e-x)

Hi Dell! :smile:

(i assume you mean /(ex - e-x))

Hint: the only difficulty is at x = 0 …

so what is the integrand approximately equal to near x = 0? :wink:
 
  • #3
my mistake, the e^e is meant to read e^x,,, don't know if that makes much of a difference

really can't find anything that helps me 1/(2x^4/3) is very close near x=0, but 1/(2x^4/3) diverges and it is bigger so it doesn't really help me,... unless i can say
lim fx/gx = 1
x->inf

can i do that since my integral is not to infinity,?? also I am not sure they ar similar when x gets vey big
 

1. What is an improper integral?

An improper integral is an integral in which one or both of the limits of integration are infinite or the integrand function is unbounded at one or more points in the interval of integration. This type of integral does not have a well-defined value and must be evaluated using other methods, such as taking limits.

2. How can I prove the convergence of an improper integral?

There are several methods for proving the convergence of an improper integral. These include the limit comparison test, the integral test, and the comparison test. Each of these methods involves comparing the given integral to a known convergent or divergent integral. If the given integral has similar behavior to the known integral, then it can be concluded that the given integral also converges or diverges.

3. Can I use the limit comparison test to prove convergence of any improper integral?

No, the limit comparison test can only be used on improper integrals with non-negative integrands. If the integrand is negative, the integral must be rewritten as the difference of two integrals, one with non-negative integrand and one with negative integrand. Then the limit comparison test can be applied to the integral with non-negative integrand.

4. Are there any other methods for proving convergence of an improper integral?

Yes, there are other methods such as the ratio test, the root test, and the alternating series test. These tests are typically used for integrals with specific types of integrands, such as power functions or alternating series.

5. Is it possible for an improper integral to converge even if the integrand is unbounded?

Yes, it is possible for an improper integral to converge even if the integrand is unbounded. This can occur if the unbounded portion of the integrand is "balanced out" by the bounded portion, or if the unbounded portion is located at one of the limits of integration. However, the convergence of such integrals must still be proven using one of the methods mentioned above.

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