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Math Help - Complex Analysis

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Complex Analysis

    Hi everyone, I have a big test tomorrow and I'm still hung up on integrating over branch cuts.

    So I was wondering how to show

     \int_0^{\infty} \frac{t^{\alpha-1}}{t+1} dt = \frac{\pi}{\sin(\pi\alpha)}
    where  0 < \alpha <1
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  2. #2
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    Quote Originally Posted by chiph588@ View Post
    Hi everyone, I have a big test tomorrow and I'm still hung up on integrating over branch cuts.

    So I was wondering how to show

     \int_0^{\infty} \frac{t^{\alpha-1}}{t+1} dt = \frac{\pi}{\sin(\pi\alpha)}
    where  0 < \alpha <1
    Consider the integral,
    \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx  \text{ where }0 < a < 1
    If we let t = \log x then we get,
    \int \limits_0^{\infty} \frac{t^{a-1}}{t+1} dt \text{ where }0 < a < 1
    Therefore, to evaluate your integral is is sufficient to evaluate the first (more convient) integral.

    Fix R>0 (and "large" i.e. large enough for all the inequalities to work out)
    let \Gamma be the rectangular contour with vertices \pm R, \pm R + 2\pi i.

    Now define the function f(z) = \frac{e^{az}}{e^z + 1} the function has a pole at z=\pi i with \text{res}(f,\pi i) = - e^{\pi a i}.

    Integrating by the residue theorem we get that,
    \int \limits_{-R}^R \frac{e^{ax}}{e^x + 1} dx + \int_0^{2\pi} \frac{e^{a(R+it)}}{e^{R+it} + 1} dt - \int \limits_{-R}^R \frac{e^{a(x+2\pi i)}}{e^{x+2\pi i} +1} dx - \int_0^{2\pi} \frac{e^{a(-R +it)}}{e^{-R+it} + 1} dt = -2\pi i e^{\pi a i }

    We will now show that the 2nd and 4th integrals (over the vertical sides) go to zero as R\to \infty.
    Look at the function inside the 2nd integral,
    \left| \frac{e^{a(R+it)}}{e^{R+it} + 1} \right| \leq \frac{|e^{a(R+it)}|}{||e^{R+it}| - |1||} = \frac{e^{aR}}{e^R - 1} \leq \frac{e^{aR}}{e^R - \frac{1}{2}e^R} = 2e^{(a-1)R}
    Look at the function inside the 4th integral,
    \left| \frac{e^{a(-R+it)}}{e^{-R+it} + 1} \right| \leq \frac{|e^{a(-R+it)}|}{||e^{-R+it}| - |1||} = \frac{e^{-aR}}{1 - e^{-R}} \leq \frac{e^{-aR}}{1 - \frac{1}{2}}  = 2e^{-aR}
    Therefore the 2nd integral satisfies,
    \left| \int_0^{2\pi} \frac{e^{a(R+it)}}{e^{R+it} + 1} dt \right| \leq 4\pi e^{(a-1)R} \to 0 \text{ since }a-1 < 0
    Therefore the 4th integral satisfies,
    \left| \int_0^{2\pi}\frac{e^{a(-R+it)}}{e^{-R+it} + 1} \right| \leq 4\pi e^{-aR} \to 0 \text{ since } -a < 0
    Look at the 3rd integral,
    \int \limits_{-R}^R \frac{e^{a(x+2\pi i)}}{e^{x+2\pi i}+1} dx = e^{2\pi a i}\int \limits_{-R}^R \frac{e^{ax}}{e^x +1} dx
    Taking the limits in the big equation with four integral above we get,
    (1 - e^{2\pi a i} ) \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx = - 2\pi i e^{\pi a i}
    Finally,
    \int \limits_{-\infty}^{\infty} \frac{e^{ax}}{e^x + 1} dx = \frac{2\pi i e^{\pi a i}}{e^{2\pi a i} - 1} = \frac{2\pi i}{e^{\pi a i} - e^{-\pi a i}} = \frac{\pi}{\sin (\pi a)}
    ----

    At this point we can derive the Euler Reflection Formula.
    Remember by the first comment we have shown that,
    \int_0^{\infty} \frac{t^{a-1}}{t+1} dt = \frac{\pi}{\sin (\pi a)}
    Let \mu = \tfrac{t}{t+1} to get,
    \int_0^{\infty} \mu ^{a-1}(1-\mu)^{-a} d\mu = \bold{B}(a,1-a) = \Gamma (a) \Gamma (1-a) = \frac{\pi}{\sin (\pi a)} \text{ for }0 < a < 1
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  3. #3
    Moo
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    Hello,

    \int_0^{\infty} \frac{t^{-\alpha}}{1+t} ~ dt=\int_0^{\infty} \frac{1}{t^\alpha (1+t)} ~ dt


    Lemma : (actually, I only know the French naming of this...)

    2i \pi \sum_{w \in \mathbb{C} \backslash \mathbb{R}_+} \text{Res}_w \left(\frac{R(z)}{z^\alpha}\right)=\left(1-e^{-2i \pi \alpha}\right) \int_0^\infty \frac{R(x)}{x^\alpha} ~ dx
    where R is a rational fraction without poles on the real axis x \geq 0 and R(x) \stackrel{|x|\to \infty}{\sim} \frac{1}{x^n},~ n \geq 1. 0<\alpha<1

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    This can be proved by considering the contour integral :
    \oint \frac{R(z)}{z^\alpha} ~ dz over this contour :

    Complex Analysis-residu.jpg

    as e \to 0,~ h \to 0,~ R \to \infty
    Let f(z)=\frac{R(z)}{z^\alpha}=\frac{R(z)}{e^{\alpha \log(z)}}



    \bullet \quad When t goes to infinity, |f(t)| \leq \frac{c}{|z|^{1-\alpha}}. Then \left|\int_\Gamma f(z) ~ dz\right| \leq \frac{c}{R^{1+\alpha}} \times R=\frac{c}{R^\alpha} \stackrel{R \to \infty}{\longrightarrow} 0

    \bullet \quad |R(z)| \leq \frac{c}{|z|^\alpha}. When z is near from 0, we thus have \left|\int_\gamma f(z) ~ dz \right| \leq c' \epsilon^{1-\alpha} \stackrel{\epsilon \to 0}{\longrightarrow} 0

    \bullet \quad \int_{\Gamma_1} f(z) ~ dz \to \int_0^{\infty} \frac{R(x)}{x^\alpha} ~ dx=I

    \bullet \quad \int_{\Gamma_2} f(z) ~ dz \to \int_0^{\infty} \frac{R(x)e^{-2i \pi \alpha}}{x^\alpha} ~ dx=e^{-2i \pi \alpha} I
    The 2i \pi \alpha comes from the fact that :
    z \to x,~ \Im z>0,~ \log z \to \ln(x) \implies z^\alpha \to x^\alpha
    But z \to x, ~ \Im z<0,~ \log z \to \ln(x)+2i \pi \implies z^\alpha \to e^{2i \pi \alpha} x^\alpha



    Hence, by the residue theorem, we have :
    I-e^{-2i \pi \alpha} I=\int_{\Gamma \cup \gamma \cup \Gamma_1 \cup \Gamma_2} f(z) ~ dz=2i \pi \sum_{w \in \mathbb{C} \backslash \mathbb{R}_+} \text{Res}_w \left(\frac{R(z)}{z^\alpha}\right)
    \blacksquare

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    There is only one pole over \mathbb{C} \backslash \mathbb{R}_+ : t=-1
    \text{Res}_{-1} ~ \frac{1}{t^\alpha(1+t)}=\frac{1}{(-1)^\alpha}

    We have \log(-1)=i \pi
    Hence (-1)^\alpha=e^{i \pi \alpha}

    By the above lemma, we get :
    2i \pi \cdot e^{-i \pi \alpha}=(1-e^{-2 i \pi \alpha}) I

    Hence :
    \frac{I}{\pi}=\frac{2i}{e^{i \pi \alpha}-e^{-i \pi \alpha}}=\frac{1}{\sin(\pi \alpha)}


    \boxed{I=\frac{\pi}{\sin(\pi \alpha)}}




    Okay, there may be mistakes and I'm sure there are unclear things, but one can just remember the formula of the lemma. Sorry for the very long message
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  4. #4
    Moo
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    If you're interested, I found completely by chance a wikipedia version of what I did : Mellin transform - Wikipedia, the free encyclopedia
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