Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Tricky Integrals

  1. Jul 5, 2005 #1
    All of these integrals have lower bounds of 0 and upper bounds of infinity:

    Problems 1 and 2 just require me to determine whether it converges or diverges. 3 and 4 actually require a value.

    1) e^(-x) * sqrt(x)
    2) [tex]\frac{x*arctan(x)}{(1+x^4)^(1/3)}[/tex]
    3) e^(-x) / sqrt(x)
    4) x^2 * e^[-(x^2)]

    I tried to use integration by parts but I went in circles

    [Edit:] number 2 should have (1+x^4)^(1/3) in the denominator
     
    Last edited: Jul 5, 2005
  2. jcsd
  3. Jul 5, 2005 #2

    lurflurf

    User Avatar
    Homework Helper

    Do you know the integral on that set for exp(-x^2)? 1,3,4 can be written in terms of it. If i am reading 2 right as (x*Arctan(x))/(1+x^4)^(1/3)~x^(-1/3) for large x and hence diverges.
     
    Last edited: Jul 5, 2005
  4. Jul 5, 2005 #3
    I am given that e^(-x^2) = 0.5*sqrt(pi), but I couldn't get the integrals into that form through integration by parts - am I doing something wrong?

    Thanks
     
  5. Jul 5, 2005 #4

    lurflurf

    User Avatar
    Homework Helper

    For 1,3 change variable u^2=x you will get 4 back from one of them and the given integral from the others. for 4 and the one that becomes like it intgrate by parts differentiate x and integrat x exp(-x^2). Again 2 (perhaps you mistyped it?) diverges.
     
  6. Jul 5, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Are you sure about the second?

    [tex] \int_{0}^{+\infty} \frac{x \arctan x}{\sqrt[3]{\left(1+x^{4}\right)}} \ dx [/tex]

    Maple cannot do it and neither Mathematica.

    Daniel.
     
  7. Jul 5, 2005 #6
    That's the correct integral, it must diverge.
     
  8. Jul 5, 2005 #7
    These are the answers that I got

    1) Converges (to 0.5*sqrt (Pi))
    2) Diverges
    3) Converges to sqrt (Pi)
    4) Converges to 0.25 * sqrt (Pi)

    Do these look right?

    Thanks
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook