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Math Help - [SOLVED] Calculating an integral via complex analysis, doubt/misunderstanding

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Calculating an integral via complex analysis, doubt/misunderstanding

    I'm having a textbook at hand (Complex Analysis by Sakarchi, Stein) and I don't understand something.
    At page 78, he wants to prove that \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi.
    He says that f(z)=\frac{1}{z^2+1} has 2 simple poles, one at -i and the other at i. Until now everything's fine.
    Now in order to calculate the integral, he chose a contour \gamma _R where this curve is the half circle with radius R, centered at the origin in the upper half plane.
    He then says that the residue of f at 1 is \frac{1}{2i}, I still follow him.
    Now he says that \int _{\gamma _R} f(z)dz=\frac{2 \pi i}{2i}=\pi. That's fine since the pole at 1 is enclosed into the curve \gamma _R.
    Now this is where I don't understand.
    He goes on to say "let C be the large half circle of radius R. We see that \left |  \int _ C f(z)dz   \right | \leq \pi R \frac{B}{R^2} \leq \frac{M}{R}. The integral goes to 0 as R \to \infty. Therefore in the limit we find \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi ".
    ------------------------------------------------------

    1)He never defined B and M (at least in the previous pages). I guess that they are arbitrary.
    2)I don't see the necessity of finding a upper bound of the integral. (Why did he did so?)
    3)He never found a curve enclosing the other pole! So how can the integral whose limits are -\infty and \infty be calculated via the residue theorem for a single pole while the function has 2 poles? Is it because Sakarchi's curve encloses all real numbers when R tends to + \infty?
    If so then I could simply chose a curve enclosing all real numbers when I take the limit, but not enclosing any poles, and the contour integral would be worth 0 according to Cauchy's integral theorem. Which makes no sense since the integral is worth \pi.

    Thanks for reading and helping.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I'm having a textbook at hand (Complex Analysis by Sakarchi, Stein) and I don't understand something.
    At page 78, he wants to prove that \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi.
    He says that f(z)=\frac{1}{z^2+1} has 2 simple poles, one at -i and the other at i. Until now everything's fine.
    Now in order to calculate the integral, he chose a contour \gamma _R where this curve is the half circle with radius R, centered at the origin in the upper half plane.
    He then says that the residue of f at 1 is \frac{1}{2i}, I still follow him.
    Now he says that \int _{\gamma _R} f(z)dz=\frac{2 \pi i}{2i}=\pi. That's fine since the pole at 1 is enclosed into the curve \gamma _R.
    Now this is where I don't understand.
    He goes on to say "let C be the large half circle of radius R. We see that \left | \int _ C f(z)dz \right | \leq \pi R \frac{B}{R^2} \leq \frac{M}{R}. The integral goes to 0 as R \to \infty. Therefore in the limit we find \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi ".
    ------------------------------------------------------

    1)He never defined B and M (at least in the previous pages). I guess that they are arbitrary.
    2)I don't see the necessity of finding a upper bound of the integral. (Why did he did so?)
    3)He never found a curve enclosing the other pole! So how can the integral whose limits are -\infty and \infty be calculated via the residue theorem for a single pole while the function has 2 poles? Is it because Sakarchi's curve encloses all real numbers when R tends to + \infty?
    If so then I could simply chose a curve enclosing all real numbers when I take the limit, but not enclosing any poles, and the contour integral would be worth 0 according to Cauchy's integral theorem. Which makes no sense since the integral is worth \pi.

    Thanks for reading and helping.
    Note that

    |z|^2=|z^2|=|z^2+1-1| \le |z^2+1|+1

    This implies that

    |z|^2-1 \le |z^2+1|

    Since we enclosed the pole at z=i we know the radius of the arc R > 1 \implies |z|^2 > 1

    This gives us the inequality

    \left| \frac{1}{z^2+1} \right| \le \left| \frac{1}{z^2-1}\right| \le \frac{1}{R^2}

    This gives us an upper bound on the valus of the integrand. So we know the value of the integral cannot be larger than its maximum multiplied by the lenth of the curve of integration.

    \left| \int_{\text{semicircle}} \frac{1}{z^2+1}dz\right|< \left|\frac{1}{R^2}dz \right| < \frac{1}{R^2}\pi R=\frac{\pi}{R}

    So the integral looks like

    \pi= \lim_{R \to \infty} \int_{-R}^{R} \frac{1}{1+x^2}dx+\int_{\text{arc}}\frac{1}{z^2+1}  dz

    Becuase of the bound we know that the 2nd integral goes to zero as R goes to infitity so we get

    \pi= \int_{-\infty}^{\infty} \frac{1}{1+x^2}dx

    As far as the other pole goes you don't need it

    The residue theorem states the values of an integral on any closed curve is the sum of the residues inside the curve. The poles outside do not affect the value of the integral.

    I hope this helps.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks for the reply, it does help.
    However I follow you until you wrote
    So the integral looks like

    How did you reach this? What do you mean by "arc"?


    As far as the other pole goes you don't need it

    The residue theorem states the values of an integral on any closed curve is the sum of the residues inside the curve. The poles outside do not affect the value of the integral.
    Ok. You're right. But I don't understand why I can't chose an arbitrary curve which doesn't enclose any pole, so that the line integral would be worth 0.

    Also, why couldn't I chose a curve enclosing both -i and i? I could have chosen the entire circle, and the integral would have been worth another value, not \pi.

    I know I'm totally missing the important points of the subject. I don't get it as of now.
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    we want to find the value of

    \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx

    let

    \gamma_1(t)=-R+2Rt, 0 \le t \le 1 and

    \gamma_2(t)=Re^{i t}, 0 \le t \le \pi, R > 1

    The union of these two curves give a closed path enclosing the the pole z=i

    So now the complex integral looks like
    \int_{\gamma_1 \cup \gamma_2}\frac{dz}{z^2+1}=\int_{\gamma_1}\frac{1}{  z^2+1}dz+\int_{\gamma_2}\frac{1}{z^2+1}dz

    By the residue theorem we know the integral on the left is \pi

    \pi=\int_{\gamma_1}\frac{1}{z^2+1}dz+\int_{\gamma_  2}\frac{1}{z^2+1}dz

    Using the paramterization of \gamma_1 we get that the first integral looks like

    \pi =\int_{-R}^{R}\frac{1}{t^2+1}dt+\int_{\gamma_2}\frac{1}{z^  2+1}dz

    Now using the bound I found in the previous post we get

    \pi =\int_{-R}^{R}\frac{1}{t^2+1}dt+\frac{\pi}{R}

    Now as we let R \to \infty

    \pi =\int_{-\infty}^{\infty}\frac{1}{t^2+1}dt+0

    The reason we want this path is we need to paramterize the real line what I called gamma 1 and then close this path so we could apply the residue theorem. The choice of the semicircle is becuase we could bound the function there and use the ML bound to show that the integral on the semicircle goes to zero as the radius goes to infinity. In Theory you could make a rectangle on any number of other shapes but the integrals my be hard or impossible to bound or compute.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    we want to find the value of

    \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx

    let

    \gamma_1(t)=-R+2Rt, 0 \le t \le 1 and

    \gamma_2(t)=Re^{i t}, 0 \le t \le \pi, R > 1

    The union of these two curves give a closed path enclosing the the pole z=i

    So now the complex integral looks like
    \int_{\gamma_1 \cup \gamma_2}\frac{dz}{z^2+1}=\int_{\gamma_1}\frac{1}{  z^2+1}dz+\int_{\gamma_2}\frac{1}{z^2+1}dz

    By the residue theorem we know the integral on the left is \pi

    \pi=\int_{\gamma_1}\frac{1}{z^2+1}dz+\int_{\gamma_  2}\frac{1}{z^2+1}dz

    Using the paramterization of \gamma_1 we get that the first integral looks like

    \pi =\int_{-R}^{R}\frac{1}{t^2+1}dt+\int_{\gamma_2}\frac{1}{z^  2+1}dz

    Now using the bound I found in the previous post we get

    \pi =\int_{-R}^{R}\frac{1}{t^2+1}dt+\frac{\pi}{R}

    Now as we let R \to \infty

    \pi =\int_{-\infty}^{\infty}\frac{1}{t^2+1}dt+0

    The reason we want this path is we need to paramterize the real line what I called gamma 1 and then close this path so we could apply the residue theorem. The choice of the semicircle is becuase we could bound the function there and use the ML bound to show that the integral on the semicircle goes to zero as the radius goes to infinity. In Theory you could make a rectangle on any number of other shapes but the integrals my be hard or impossible to bound or compute.
    Thanks, that was very clear.
    Now I must read more on this, and of course, do exercises.
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