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yuiop
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I this old thread it mentions that the indefinite integral of f'(x)/f(x) is log(|f(x)|)+C which means that there is some ambiguity about the sign of f(x). There does however, seem to be no ambiguity about the value of C as it always appears to be zero, but I have never seen this mentioned anywhere. I have only tested it using random real values, so I am not sure if it is generally applicable with complex numbers, etc.
I noticed this while carrying out the indefinite integral of an expression of the form ##\frac{1}{J(r)}\frac{J(r)}{dr}##. For example consider:
##J(r) = K(r) + X## where X is a constant. Take the derivative:
##\frac{dJ(r)}{dr} = \frac{d(K(r)+X)}{dr} = K'(r)##. Divide both sides by J(r):
##\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{K'(r)}{K(r)+X}##. Try to recover the original expression by taking the indefinite integral:
##\int{\left(\frac{1}{J(r)}\frac{dJ(r)}{dr}\right)} dr =\int{\left(\frac{K'(r)}{K(r)+X}\right)} dr= \log(|K(r)+X|)= \log({|J(r)|})##
In other words, when the integral is taken the original constant X is recovered and the constant of integration C is not required.
I am not a mathematician so this is not intended to be a formal proof that ##\int \left(f'(x)/f(x)\right) = \log {(|f(x)|)}## rather than ##\log{(|f(x)|)}+C##. I thought maybe some of you would like to check it out.
I noticed this while carrying out the indefinite integral of an expression of the form ##\frac{1}{J(r)}\frac{J(r)}{dr}##. For example consider:
##J(r) = K(r) + X## where X is a constant. Take the derivative:
##\frac{dJ(r)}{dr} = \frac{d(K(r)+X)}{dr} = K'(r)##. Divide both sides by J(r):
##\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{K'(r)}{K(r)+X}##. Try to recover the original expression by taking the indefinite integral:
##\int{\left(\frac{1}{J(r)}\frac{dJ(r)}{dr}\right)} dr =\int{\left(\frac{K'(r)}{K(r)+X}\right)} dr= \log(|K(r)+X|)= \log({|J(r)|})##
In other words, when the integral is taken the original constant X is recovered and the constant of integration C is not required.
I am not a mathematician so this is not intended to be a formal proof that ##\int \left(f'(x)/f(x)\right) = \log {(|f(x)|)}## rather than ##\log{(|f(x)|)}+C##. I thought maybe some of you would like to check it out.
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