Integral inconsistency with variable change

[matt_crouch]

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?

 

[axmls]

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.

 

[Svein]

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, $dw = dx$ and the integral becomes $\int(x+y)dx=\frac{x^{2}}{2}+yx+C$. Furthermore, $\frac{1}{2}w^{2}=\frac{1}{2}(x+y)^{2}=\frac{1}{2}x^{2}+\frac{2}{2}xy+\frac{1}{2}y^{2}$. Since y is constant, you can absorb the y² part into the integration constant C.

 

[matt_crouch]

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, $dw = dx$ and the integral becomes $\int(x+y)dx=\frac{x^{2}}{2}+yx+C$. Furthermore, $\frac{1}{2}w^{2}=\frac{1}{2}(x+y)^{2}=\frac{1}{2}x^{2}+\frac{2}{2}xy+\frac{1}{2}y^{2}$. Since y is constant, you can absorb the y² part into the integration constant C.

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?

 

[Svein]

However you can do the substitution for example in complex analysis where u=x+iyu=x+iy and v=x−iyv=x-iy with the differentials becoming du=dx+idydu=dx+idy and dv=dx−idydv=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?

It is a bit more complicated than that.
A complex integral is always taken along a specified path, which means that instead of writing $\int_{a}^{b}f(x)dx$, you have to write $\int_{\gamma}f(z)dz$ in the complex case. Also $\gamma=\gamma(t)$, with t∈[0, 1] and $\gamma(0)=a, \gamma(1)=b$. The implication is that complex integrals tend to end up as $\int_{0}^{1}f(\gamma(t))\frac{d\gamma}{dt}dt$ - which again is a one-dimensional integral.

 

[matt_crouch]

It is a bit more complicated than that.
A complex integral is always taken along a specified path, which means that instead of writing $\int_{a}^{b}f(x)dx$, you have to write $\int_{\gamma}f(z)dz$ in the complex case. Also $\gamma=\gamma(t)$, with t∈[0, 1] and $\gamma(0)=a, \gamma(1)=b$. The implication is that complex integrals tend to end up as $\int_{0}^{1}f(\gamma(t))\frac{d\gamma}{dt}dt$ - which again is a one-dimensional integral.

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

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