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Math Help - Why does the integrand in complex integral need to be continuous ?

  1. #1
    Aki
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    Question Why does the integrand in complex integral need to be continuous ?

    I found in textbooks of complex analysis that the integral of a complex function f(z)

    <br />
I = \int_C f(z) dz<br />

    is defined only for continuous functions.
    I know that the integral of a complex function is reduced to curvilinear integrals of real functions as follows

    <br />
I = \int_C(u dx - v dy) + i \int_C(u dy + v dx) <br />

    ,where

    z=x+iy,   f(z)=u(x,y)+iv(x,y).

    And, I understand that in curvilinear integral the integrand doesn't need to be continuous.
    If it is true, why does the integrand in complex integral need to be continuous ?
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  2. #2
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    It can be shown that if a function of a complex variable is differentiable (in some neighborhood, not just at a point), then it is analytic, which immediately implies it has derivatives of all orders, which, finally, means that its deerivative is differentiable, and so continuous.

    If, by F(z)= \int f(z)dz, we mean the function whose derivative, in some region, is 0, the F must have a derivative, which implies it is analytic, which implies it derivative, f(z), is continuous.
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    Quote Originally Posted by Aki View Post
    I found in textbooks of complex analysis that the integral of a complex function f(z)

    <br />
I = \int_C f(z) dz<br />

    is defined only for continuous functions.
    why does the integrand in complex integral need to be continuous ?
    I don't think it has to be. The integral of a complex-valued function is still the limit of a Riemann sum. No different than for real functions. So take for example:

    f(z)=\begin{cases}z^2 & |z|\leq 2 \\<br />
                                  z^3 & |z|>2<br />
\end{cases}<br />

    Then:

    \int_0^3 f(z)dz

    still exists.
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  4. #4
    Aki
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    Thanks, HallsofIvy!

    Quote Originally Posted by HallsofIvy View Post
    If, by F(z)= \int f(z)dz, we mean the function whose derivative, in some region, is 0, the F must have a derivative, which implies it is analytic, which implies it derivative, f(z), is continuous.
    I don't understand why the function's derivative is 0 in some region.
    Could you explain this to me ?
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  5. #5
    Aki
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    Thank you for helping me, shawsend !

    I see.
    If f(z) has discontinuities at the points {p_1 ,\cdots,p_n} on C (integral path), you can still integrate it by cutting C at {p_1 ,\cdots,p_n} into {C_1 ,\cdots,C_{n+1}} and dividing the integral as follows:

    <br />
I=\int_{C_1}f(z)dz +\cdots+\int_{C_{n+1}}f(z)dz.<br />

    But, I'm not sure.
    Is it the same thing to say that the Reimann sum

    <br />
\Sigma_\Delta=\Sigma f(\zeta_i)(z_i -z_{i-1}),<br />

    where z_i's are points on C and \zeta_i is a point on the arc (z_i z_{i-1}) , converges to 0 as \max_i |z_i -z_{i-1}| goes to 0 regardless of the selection of the points z_i's ?

    PS
    Just a small thing.
    As the function you suggested is not an analytic function,
    you have to specify the path of integration.
    Last edited by Aki; July 25th 2010 at 10:26 PM.
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  6. #6
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    That was a typo (and an unusually bad one!) I meant to say
    "If, by F(z)= \int f(z)dz, we mean the function whose derivative, in some region, is f(z)". That was specifically to avoid examples such as shawsend's.
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  7. #7
    Aki
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    Quote Originally Posted by HallsofIvy View Post
    That was a typo (and an unusually bad one!) I meant to say
    "If, by F(z)= \int f(z)dz, we mean the function whose derivative, in some region, is f(z)". That was specifically to avoid examples such as shawsend's.
    Thanks.

    Now, I understand.
    If it is an indefinite integral, the integrand must be analytic, therefore continuous, as only analytic functions have their primitives.
    What about definite integrals ?
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  8. #8
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    ??? Definite integrals are NUMBERS. "Analytic", "continuous", "differentiable", etc. apply to functions, not numbers.
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  9. #9
    Aki
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    Quote Originally Posted by HallsofIvy View Post
    ??? Definite integrals are NUMBERS. "Analytic", "continuous", "differentiable", etc. apply to functions, not numbers.
    Maybe, my bad English is causing a confusion.
    Let me restate my question.

    The Reimann sum of a real function f(x) over the closed inverbal [a,b]

    <br />
\Sigma_\Delta=\Sigma_{i=0}^n f(\xi_i)(x_i -x_{i-1}),<br />

    where a=x_0<x_1<\cdots<x_{n+1}=b and x_{i-1}<\xi_i<x_i, converges to a limit regardless of the selection of \{x_1,\cdots,x_n\} when \delta=\max_{i=0}^n |x_i -x_{i-1}| goes to zero if f(x) is bounded over [a,b].
    It is still true even f(x) has a finite number of discontinuities over the interval, since the contributions to \Sigma_\Delta from the discontinuities tends to zero when \delta approaches zero.

    On the other hand, the definite integral of a complex function f(z)

    <br />
I = \int_C f(z) dz,<br />

    where C is a smooth curve on the complex plane, can be reduced to real curvilinear integrals as follows:

    <br />
I = \int_C(u dx - v dy) + i \int_C(u dy + v dx),<br />

    where z=x+iy, f(z)=u(x,y)+iv(x,y). Then, it seems to me that it is possible to define the definite integral even if f(z) has, at least, a finite number of discontinuities on C as in the integrals of real functions. But, the integral of complex functions is defined only for continuous functions in the textbooks I saw. Perhaps, I am misunderstanding something. Could you point it out ?
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  10. #10
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    Quote Originally Posted by Aki View Post
    Maybe, my bad English is causing a confusion.
    Let me restate my question.

    The Reimann sum of a real function f(x) over the closed inverbal [a,b]

    <br />
\Sigma_\Delta=\Sigma_{i=0}^n f(\xi_i)(x_i -x_{i-1}),<br />

    where a=x_0<x_1<\cdots<x_{n+1}=b and x_{i-1}<\xi_i<x_i, converges to a limit regardless of the selection of \{x_1,\cdots,x_n\} when \delta=\max_{i=0}^n |x_i -x_{i-1}| goes to zero if f(x) is bounded over [a,b].
    It is still true even f(x) has a finite number of discontinuities over the interval, since the contributions to \Sigma_\Delta from the discontinuities tends to zero when \delta approaches zero.
    I'm not an analyst but I believe that's true as long as the discontinuities are bounded and sometimes the integral exist even if the discontinuity is unbounded such as:

    \int_0^1 \log(x)dx=-1

    And keep in mind that's true even if I consider the complex-valued log function (principal value) over that contour.

    On the other hand, the definite integral of a complex function f(z)

    <br />
I = \int_C f(z) dz,<br />

    where C is a smooth curve on the complex plane, can be reduced to real curvilinear integrals as follows:

    <br />
I = \int_C(u dx - v dy) + i \int_C(u dy + v dx),<br />

    where z=x+iy, f(z)=u(x,y)+iv(x,y). Then, it seems to me that it is possible to define the definite integral even if f(z) has, at least, a finite number of discontinuities on C as in the integrals of real functions.
    Also true as long as the discontinuities are bounded and you don't have to define them in terms of real integrals. They are identical to limits of real Riemann sums except that complex numbers are being added in the sum and (complex) antiderivatives can still be used to evaluate them over regions of analyticity.

    But, the integral of complex functions is defined only for continuous functions in the textbooks I saw. Perhaps, I am misunderstanding something. Could you point it out ?
    I'm beginning to think you're getting confused about whether the integral exists as opposed to solving the integral by various techniques that require the function to be analytic such as Cauchy's Theorem and the Residue Theorem. The integral exists if the limit of the Riemann sum exists. However it cannot be evaluated singly, by the Residue Theorem or by using a single antiderivative if the function is discontinuous at points over the path of integration.
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