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Hello, can anyone tell me how I can show that this

  1. Nov 2, 2011 #1
    Hello, can anyone tell me how I can show that this diverges

    [tex]\int_{-1}^0\frac{e^x}{x+1}dx[/tex]

    I'm out of ideas
     
  2. jcsd
  3. Nov 2, 2011 #2

    lavinia

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    Re: Divergence

    When in doubt try integration by parts
     
  4. Nov 2, 2011 #3
    Re: Divergence

    Yes, of course. Thanks a lot!
     
  5. Nov 2, 2011 #4
    Re: Divergence

    No, it's not working, I really can't evaluate this integral. Is there another way to see that it diverges?
     
  6. Nov 2, 2011 #5
    Re: Divergence

    As a warm-up, why does the integral of 1/x from 0 to 1 diverge?

    It's not too different from that. Maybe use the comparison test, since you can bound e^x from below by some number greater than zero.
     
  7. Nov 2, 2011 #6
    Re: Divergence

    Because [tex]\lim_{x\to0}\frac1{x}[/tex] diverges? I'm not sure. What's the comparison test?
     
  8. Nov 2, 2011 #7

    disregardthat

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    Re: Divergence

    Find a lower bound for the integrand on [-1,0], that is, a function that is less than your integrand on [1,0]. (technically the absolute value of the integrand, but in this case it is positive)

    Hint: what is the minimal value of e^x on [-1,0] ?

    Then you can use the comparison test to see whether the integral of the lower bound diverges.
     
  9. Nov 2, 2011 #8

    lavinia

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    Re: Divergence

    Integrating by parts gives

    [tex]\frac{e^x}{x+1}|^{0}_{-1}+ \int_{-1}^0\frac{e^x}{(x+1)^2}dx[/tex]

    The first term diverges. If the second term is finite then the whole thing diverges. What if the second term diverges?
     
    Last edited: Nov 2, 2011
  10. Nov 6, 2011 #9
    Re: Divergence

    If one part diverges, then the whole thing diverges
     
  11. Nov 6, 2011 #10

    lurflurf

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    Re: Divergence

    ^No a sum of two divergent terms can converge such as (1/x-1/x) which converges even though each part diverges. For this problem show that
    [tex]\int_{-1}^0\frac{e^x+a}{x+1}dx[/tex]
    can converge only when a =-e-1
    and in particular not when a=0
     
  12. Nov 7, 2011 #11
    Re: Divergence

    Because ln(0) is slightly undefined?
     
  13. Nov 11, 2011 #12
    Re: Divergence

    integral -1 to 0 (e^x)/1+x>e^-2[integral -1 to 0 1/(1+x)]


    The second term diverges so first term diverges.
     
  14. Nov 11, 2011 #13

    lavinia

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    Re: Divergence

    The argument i was making was not that. The first term, is easily seen to diverge since the numerator is finite and the denominator blows up at -1. It diverges to +infinity. If the second term is finite then the sum must diverge. But since the second term is positive, if it diverges, then it also diverges to plus infinity so the sum of the two must diverge to plus infinity.
     
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