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Can i do this to prove convergence of improper integral

  1. Apr 12, 2009 #1
    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 doesnt 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 doesnt work. what can i do from here??

    if not how would you solve this?
     
  2. jcsd
  3. Apr 12, 2009 #2

    tiny-tim

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    Science Advisor
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    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:
     
  4. Apr 12, 2009 #3
    my mistake, the e^e is meant to read e^x,,, dont know if that makes much of a difference

    really cant 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 doesnt 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 im not sure they ar similar when x gets vey big
     
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