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Integral inconsistency with variable change

  1. Jun 22, 2015 #1
    so

    ## \int wdw = \frac{1}{2}w^{2} ##

    now if w=x+y,

    ## \int (x+y)(dx+dw)= \int xdx + \int ydx + \int xdy + \int ydy ##

    which can be evaluated and gives

    ##\int(x+y)(dx+dy)=\frac{1}{2}x^{2}+2xy+\frac{1}{2}y^{2}##

    but

    ##\frac{1}{2}x^{2}+2xy+\frac{1}{2}y^{2} \neq \frac{1}{2}w^{2}##

    can someone explain why?
     
    Last edited: Jun 22, 2015
  2. jcsd
  3. Jun 22, 2015 #2
    There's a hidden danger behind treating differentials as actual variable to manipulate, and so it should be generally avoided except in cases in which one knows why what they're doing is mathematically sound.
     
  4. Jun 22, 2015 #3

    Svein

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    You are turning a standard integral into a double integral without any justification. Your substitution w = x + y is valid when either x or y is a constant. Thus, assuming y to be a constant, [itex]dw = dx [/itex] and the integral becomes [itex]\int(x+y)dx=\frac{x^{2}}{2}+yx+C [/itex]. Furthermore, [itex] \frac{1}{2}w^{2}=\frac{1}{2}(x+y)^{2}=\frac{1}{2}x^{2}+\frac{2}{2}xy+\frac{1}{2}y^{2}[/itex]. Since y is constant, you can absorb the y2 part into the integration constant C.
     
  5. Jun 24, 2015 #4
    However you can do the substitution for example in complex analysis where ##u=x+iy## and ##v=x-iy## with the differentials becoming ##du=dx+idy## and ##dv=dx-idy##, Is this substitution and hence the differentials valid because of the imaginary numbers? and hence when you integrate you can separate real and imaginary parts?
     
  6. Jun 25, 2015 #5

    Svein

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    It is a bit more complicated than that.
    A complex integral is always taken along a specified path, which means that instead of writing [itex]\int_{a}^{b}f(x)dx [/itex], you have to write [itex]\int_{\gamma}f(z)dz [/itex] in the complex case. Also [itex]\gamma=\gamma(t) [/itex], with t∈[0, 1] and [itex] \gamma(0)=a, \gamma(1)=b[/itex]. The implication is that complex integrals tend to end up as [itex]\int_{0}^{1}f(\gamma(t))\frac{d\gamma}{dt}dt [/itex] - which again is a one-dimensional integral.
     
  7. Jun 28, 2015 #6
    Ok thanks for your response. With regard to the first integral in the first post, you actually can show that they are both equivalent,

    ##\int (x+y)(dx+dy)=\int xdx +\int xdy+ \int ydx +\int ydy##
    from the relationship
    ##dw=dx+dy##
    we also get
    ##dx=dw-dy##, ##dy=dw-dx##

    if you substitute those into the cross variable terms

    ##\int xdx+\int ydy +\int x(dw-dx)+\int y(dw-dy)##,
    ##\int dx+ \int ydy +\int xdw -\int xdx -\int ydy +\int ydw##

    This will simplify to

    ##\int (x+y)dw=\int wdw=\frac{1}{2}w^{2}##
     
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