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Contour integral methods

  1. Aug 18, 2009 #1
    Hi there!

    I am trying to prove the following 2 identities using complex analysis methods and contour integration and I'm really stuck on defining the integration paths.

    [tex]\int_{0}^{1}\frac{\log(x+1)}{x^2+1}d x=\frac{\pi\log2}{8}[/tex]

    [tex]\int_{0}^{\infty}\frac{x^3}{e^x-1}d x=\frac{\pi^2}{15}[/tex]

    It's interesting that the first integral is proper in both limits of integration, whereas the second - improper in both limits.

    I am familiar with a proof of the second identity using series, uniform convergence and the gamma function. This proof also verifies the following generalisation of the second identity:

    [tex]\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}d x=\Gamma(s)\zeta(s)[/tex]

    But I'm looking for a way in the complex plain.

    Any help or hints are much appreciated!

    Regards, Marin
  2. jcsd
  3. Aug 19, 2009 #2


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  4. Aug 19, 2009 #3
    yeah, I am familiar with most of the methods, yet as already mentioned I cannot set the correct path (contour).

    This is because of the integrand functions; they don't fit the conditions precicely...
  5. Aug 19, 2009 #4
    Ok here's my approach on the second integral, I hope some of you can confirm or correct it :)

    We define the following contour: Set the cut for the log to be the positive real axis. We start at the origin, in fact at a circle of radius epsilon around the origin (epsilon will tend towards 0), then moving straight, parallel and slightly above to the real axis ([tex]z=x+i\delta[/tex]) up to R, the radius of the second circle round the origin (R will tend to infty). So, we go round till we almsot reach the positive real axis ([tex]z=x-i\delta[/tex]), then again towards the first circle.(sorry, don't know how to make a sketch with tex)

    The defined curve is closed and rectifiable (Jordan curve)

    We are going to apply the residue thm. on the function

    [tex]f(z)=\frac{z^3\log z}{e^z-1}[/tex]

    The above set contour could be separated in four different paths:

    [tex]\gamma_1(x):=x+i\delta[/tex], [tex]x\in[\varepsilon,R][/tex]
    [tex]\gamma_2(x):=Re^{ix}, x\in[0, 2\pi][/tex]
    [tex]\gamma_3(x):=x-i\delta[/tex], [tex]x\in[R,\varepsilon][/tex]
    [tex]\gamma_4(x):=\varepsilon e^{-ix}, x\in[0, 2\pi][/tex]

    by the residue thm. for the zeros of the denominator of f (i.e. c_k) we have:

    [tex]2\pi i\displaystyle{\sum_k Res_{c_k} f(z)}=\displaystyle{\oint_{\gamma}f(z)dz}=\displaystyle{\int_{\gamma_1}f(z)dz}+\displaystyle{\int_{\gamma_2}f(z)dz}+\displaystyle{\int_{\gamma_3}f(z)dz}+\displaystyle{\int_{\gamma_4}f(z)dz}[/tex]

    We will proove that the integrals of gamma_2 and gamma_4 (I will use Gamma for both) vanish for R->oo and epsilon->0:


    By the estimation lemma we get:

    [tex]\displaystyle{|\oint_{\Gamma}f(z)dz|}=\displaystyle{|\oint_{\Gamma}\frac{z^3\log z}{e^z-1}dz|}\leq 2\pi R||\frac{z^3\log z}{e^z-1}||_{\Gamma}[/tex]

    where, ||-|| denotes the supremum norm on Gamma

    Now, consider the following inequalities:

    [tex]||e^z-1||_{\Gamma}\geq ||e^z||_{\Gamma}-1= ||e^{Re^{\pm ix}}||-1\geq e^R-1[/tex]

    [tex]||\log z||_{\Gamma}=||\log Re^{\pm ix}||=||\log R\pm 2\pi i||\leq \log R+2\pi[/tex] (the - sign considers the curve gamma_3 but the result remains unchanged)

    [tex]||z^3||_{\Gamma}=||R^3e^{\pm 3ix}||=R^3[/tex]

    so, altogether we get:

    [tex]\displaystyle{|\oint_{\Gamma}\frac{z^3\log z}{e^z-1}dz|}\leq 2\pi R||\frac{z^3\log z}{e^z-1}||_{\Gamma}\leq 2\pi\frac{R^4(\log R+2\pi)}{e^R-1}=2\pi\frac{{\varepsilon}^4(\log\varepsilon+2\pi)}{e^{\varepsilon}-1}[/tex]

    Now, using l'Hospital's rule the proof of the statement is finished:

    [tex]2\pi\frac{R^4(\log R+2\pi)}{e^R-1}\longrightarrow^{R\rightarrow\infty}0[/tex]

    [tex]2\pi\frac{{\varepsilon}^4(\log\varepsilon+2\pi)}{e^{\varepsilon}-1}\longrightarrow^{\varepsilon\rightarrow 0} 0[/tex]

    Now, consider the integrals for gamma_1 and gamma_3:

    [tex]\displaystyle{\int_{\gamma_1}f(z)dz}+\displaystyle{\int_{\gamma_3}f(z)dz}=\displaystyle{\int_{\varepsilon}^R\frac{(x+i\delta)^3\log (x+i\delta)}{e^{x+i\delta}-1}}+\displaystyle{\int^{\varepsilon}_R\frac{(x-i\delta)^3\log (x-i\delta)}{e^{x-i\delta}-1}}[/tex]

    now, letting delta towards 0, and considering the plain cut we obtain

    [tex]\displaystyle{\int_{\varepsilon}^R\frac{x^3\log x}{e^{x}-1}}-\displaystyle{\int_{\varepsilon}^R\frac{x^3\log (x+2\pi i)}{e^{x}-1}}=-2\pi i\displaystyle{\int_{\varepsilon}^R\frac{x^3}{e^{x}-1}}[/tex]

    which, after letting epsilon towards 0 and R towards infty, becomes the sought integral.

    [tex]-2\pi i\displaystyle{\int_0^{\infty}\frac{x^3}{e^{x}-1}}=2\pi i\displaystyle{\sum_k Res_{c_k} f(z)}[/tex]

    As the denominator has infinite many zeroes, but discrete and countable, the residue sum becomes a series, also to be computed.

    All we have to do now is to calculate the residue of f(z) for the zeroes of the denominator, i.e. [tex]c_k=2\pi ik,k\in N[/tex]

    From the l'Hospital's rule we conclude that the singularities are poles and have all order 3 (for k=0 the singularity is removable), so

    [tex]Res_{2\pi ik} f(z)=\displaystyle{\lim_{z\rightarrow 2\pi ik}\frac{d^2}{dz^2}\frac{(z-2\pi ik)^3z^3\log z}{e^z-1}}[/tex]

    Now, I am not sure about the residue, but to my calculations it should be

    [tex]Res_{2\pi ik} f(z)=-8ik^3\pi^3\log(2\pi ik)[/tex]

    which does not prove the initial integral identity. That's why I suspect there is a mistake in my calculations. If you find it, I would be glad to know.

    I'm also still looking for the path suitable for the first integral.

    Best regards, Marin
    Last edited: Aug 20, 2009
  6. Aug 21, 2009 #5
    Do you have any reason to think that the first integral is amenable to complex-analytic approaches? As far as I know, the integral is suited for a straight up substitution approach (there are multiple substitutions that will get you the right answer). You can also do the problem elegantly by differentiating under the integral sign. You may obtain an ugly solution through series expansion, but the complex-analytic approach, if one even exists, is probably not going to be elegant at all (which to me seems to be the first reason to use contour integration at all).
  7. Aug 23, 2009 #6
    what do you mean by differentiating under the integral sign?

    Here's how I interpret it:

    define a function F(t) (using integral) such that t appears as a parameter in the integrand and for a certain value of t , say t_0 the integral becomes the sought one. Ok, by differentiating you could obtain an easier to solve integral for the derivative of F, i.e. F'(t). Then we have to integrate F' over t and in the end substitute the value t_0, but how do we know how to correctly set the arbitrary constant or the limits of integration?


    we seek a function G(t) (defined as a proper integral from 0 do 1)such that its derivative G'(t) for a ceratain value of t, say t_0 is the sought integral. This method gets rid of the problem with the arbitrary constant, although I don't know how to generally search for such G?

    Any opinions for the contour integration of the second integral are much appreciated

    PS: I forgot the 1/2! factor in the residue term
  8. Aug 23, 2009 #7
  9. Aug 24, 2009 #8
    thanks, that's what I needen :)
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