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I Question about Complex limits of definite integrals

  1. Jan 30, 2017 #1
    Hi, I see a formula of gamma function and i have a question.
    (1) $$\Gamma (s) = \int_{0}^{\infty } e^{-x}\, x^{s-1} dx$$
    (2) $$ x=a\, n^{p} \rightarrow dx=ap\, n^{p-1}dn$$
    (3) $$\frac{\Gamma (s)}{pa^{s}} = \int_{0}^{\infty } e^{-an^{p}}\, n^{ps-1} dn$$

    i understand the formula but why works for complex values of a?
    i mean the substitution in (2) is valid for ##a > 0## and ##p>0##
    Because otherwise the upper limit would be ##\pm \infty \, i ## or ## \infty \ \pm \infty \, i ##
    Depending on the value of a.

  2. jcsd
  3. Jan 30, 2017 #2
    Let's define for all [itex]-\frac{\pi}{2}<\theta<\frac{\pi}{2}[/itex] a line
    \gamma_{\theta} = \{\tau e^{i\theta}\;|\; \tau\in\mathbb{R},\;0\leq \tau\}
    Do you know how to prove the formula
    \int\limits_0^{\infty} e^{-x}x^{s-1}dx = \int\limits_{\gamma_{\theta}} e^{-z}z^{s-1}dz
  4. Jan 31, 2017 #3
  5. Jan 31, 2017 #4
    By definition of an integral over a parameterized path the integral on right is the same as
    \int\limits_{\gamma_{\theta}} e^{-z} z^{s-1} dz = \int\limits_0^{\infty} e^{-\tau e^{i\theta}} (\tau e^{i\theta})^{s-1} e^{i\theta} d\tau
    which is roughly the same thing you were asking about. For this reason it would be better to speak about modifying the integration path than performing "a complex change of variables". There is some theorem that states how the change of variable works, but the theorem assumes that the change of variable happens with real parameters, so if you try to apply "a complex change of variables", you are not obeying the assumptions of the theorem.

    Next question is that do you know what Cauchy's integral formula and residue theorem are? If not, you'll have to study complex analysis to learn about those. The formula I asked about can be proven with Cauchy's integral formula (or residue theorem).
  6. Jan 31, 2017 #5
    I studied something on the subject But I do not quite understand it, I know that a complex integral Can be "separated" in the cauchy main value and on the path that follows the integral, so the trick Was to show that the path of the integral is 0, so the principal value and the complex integral are the same, the reason why the change of variable is right is because something like that?
  7. Jan 31, 2017 #6
    No, this is not related to principal value stuff. By Cauchy's integral formula and residue theorem I meant these:



    I see that mentioning the Cauchy's integral formula was slightly confusing because there is no pole in sight in your problem, but anyway, of course Cauchy's integral formula is related in the sense that the residue theorem is based on it. In your problem we are going to apply the residue theorem in the special case that no poles are present, the value of the integral over a closed loop is going to be zero.

    Was there a special name for a result that the value of a complex integral is zero, if there are no poles enclosed? I don't remember it anymore, if there was.

    The formula
    z\mapsto e^{-z}e^{(s-1)\log(z)}
    defines a complex analytic function on the set [itex]\mathbb{C}\setminus\;]-\infty,0][/itex] when we use the branch of logarithm that is continuous elsewhere but not on the line [itex]]-\infty,0][/itex].

    For simplicity, now we'll assume [itex]0<\theta<\frac{\pi}{2}[/itex]. We'll define a loop that consist of three pieces by setting
    \gamma^{\textrm{arc}}_{\theta,R} = \big\{Re^{i\varphi}\;\big|\; 0\leq \varphi\leq\theta\big\}
    \gamma^{\textrm{line}}_{\theta,R} = \big\{\tau e^{i\theta}\;\big|\; 0\leq \tau\leq R\big\}

    The residue theorem will imply that
    \int\limits_{[0,R]\cup \gamma^{\textrm{arc}}_{\theta,R}\cup \gamma^{\textrm{line}}_{\theta,R}} e^{-z}z^{s-1}dz = 0
    because there are no poles enclosed by this loop. The integral is the same as
    \int\limits_0^R e^{-x}x^{s-1}dx + \int\limits_0^{\theta} e^{-Re^{i\varphi}}(Re^{i\varphi})^{s-1} iRe^{i\varphi} d\varphi
    - \int\limits_0^R e^{-\tau e^{i\theta}} (\tau e^{i\theta})^{s-1} e^{i\theta} d\tau

    Then you'll have to figure out a way to prove that the integral over the arc part goes to zero in the limit [itex]R\to\infty[/itex].
  8. Jan 31, 2017 #7
    thanks , even if i dont understand completely I know that the integral on a closed curve without poles is 0 so I got an idea of the general concept =D
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