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Can i do this to prove convergence of improper integral

  • Thread starter Dell
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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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
590
0
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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