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6 integrals using contours

  1. Apr 26, 2007 #1
    Hello ppl,

    I'm trying to solve these 6 improper integrals using calculus of residues.
    OK, I have actually got 7 now...

    (1) [tex]\int_{0}^{\infty} \frac{\ln(1+x)}{1+x^{2}} dx[/tex]
    PS: I already know how to solve
    [tex]\int_{0}^{\infty} \frac{\ln(x)}{1+x^{2}} dx[/tex]
    which equals 0, where ln(z) is a multiple-valued function
    in the complex domain with branch point z=0.
    But I didn't know what contour to use for (1) since
    the branch point of ln(1+z) is at z=-1. If I indent it
    at z=-1 and use a similar shaped contour to that of
    ln(z), I get the contribution from -1 to 0 in addition to
    the contribution from 0 to [tex]\infty[/tex], which
    throws me off.

    (2) [tex]\int_{0}^{\infty} \frac{\ln(1+x+x^{2})}{1+x^{2}} dx[/tex]
    PS: Again, I know how to solve
    [tex]\int_{0}^{\infty} \frac{\ln(1+x^{2})}{1+x^{2}} dx[/tex]
    which equals [tex]\pi \ln 2[/tex], but the presence of "x"
    within [tex]1+x+x^{2}[/tex] in (2) is giving me a hard time.

    (3) [tex]\int_{0}^{\infty} \frac{\ln^{3}(1+x^{2})}{1+x^{2}} dx[/tex] i.e. [tex]\int_{0}^{\infty} \frac{(\ln(1+x^{2}))^{3}}{1+x^{2}} dx[/tex]

    (4) [tex]\int_{0}^{\infty} \frac{x \ln(1+x^{2})}{1+x^{2}} dx[/tex]

    (5) [tex]\int_{0}^{\infty} \frac{\ln(1+x^{2})}{(1+x^{2}) \sqrt{4+x^{2}}} dx[/tex]

    (6) [tex]\int_{0}^{\infty} \frac{\sqrt{x} \ln{(1+x)}}{1+x^{2}} dx[/tex]
    Here, the numerator consists of product of 2 multi-valued functions
    with differing branch points within the complex domain
    [one at z=0 for [tex]\sqrt{z}[/tex], the other at z=-1 for ln(1+z)].
    How do I tackle 2 branch points and what would be the best
    contour to use here?

    (7) [tex]\int_{0}^{\infty} \frac{\sqrt{x} \sin^{-1}(1+x)}{1+x^{2}} dx[/tex]
    Again, the numerator here consists of product of 2 multi-valued functions
    (sqrt and arc sin); with one branch point at z=0 for [tex]\sqrt{z}[/tex],
    and then 2 branch cuts for [tex]\sin^{-1}(1+z)[/tex]
    (i) from [tex]-\infty[/tex] to -2, and (ii) from 0 to [tex]\infty[/tex].

    ---Nasim (nasim09021975@gmail.com)
    Last edited: Apr 26, 2007
  2. jcsd
  3. Apr 26, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Were you given these for homework or just wondered how to do them? Because the solutions I get for some of these are extremely difficult.

    My solution to number 2, which agrees with Mathematica, is 8 lines long and contains non elementary functions.
  4. Apr 26, 2007 #3
    wow, that is cool, Gib Z ! :eek: will you please let me know (to
    the best of your ability) how you solved them, at least if I can
    observe/witness/critic your way of thinking, I believe I can
    learn more of the amazing and beautiful topic of residue calculus!

    after all, I am here to learn, aren't I ? o:)
  5. Apr 26, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    Actually I didn't do them by hand, I probably couldn't. I am just telling you that a Computer Program that is very intelligent gets a very long answer.
  6. Apr 26, 2007 #5
    Mathematica often gives complicated answers with stuff like the LambertW function even if the answer can be simple.
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