The problem "Fundamental Rule of Leibniz?" greatly confused me about the constant of integration. In the question, \[f\left( x \right) =\int { \frac { 1 }{ x } } dx\quad and\] \[f\left( -1 \right) =1\] We are asked to find the value of \[f\left( 1 \right) \] According to the answer, the value cannot be found. \[f\left( x \right) =\ln { \left| x \right| } +C\] \[f\left( -1 \right) =\ln { \left| -1 \right| } +C=1\] \[This\quad gives\quad C=1\] \[Now,\quad f\left( 1 \right) =\ln { \left| 1 \right| } +C'\quad =C'\] I think that C should be equal to C' but according to the answer,\[ C\neq C'\] I am unable to understand it. Can someone please explain it.

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TopNewestI think the issue may be the lack of continuity of the function \(\dfrac{1}{x}\) at \(x = 0.\) First, I'm wondering if the initial statement of the question should in fact be

\(\displaystyle f(x) = \int_{a}^{x} \dfrac{1}{t} dt\) with \( f(-1) = 1\) for some non-zero real value \(a.\)

Then \(f(x) = \ln |x| - ln |a| \Longrightarrow f(-1) = \ln |-1| - \ln |a| = 1 \Longrightarrow \ln |a| = -1 \Longrightarrow a = \pm \dfrac{1}{e}.\)

This interpretation is still consistent with \(C = 1,\) but I have phrased the question so that \(C = -\ln |a|.\) Further, to avoid the discontinuity of \(f(x)\) at \(x = 0,\) for negative values of \(x\) we would need to make \(a = -\dfrac{1}{e},\) and for positive values of \(x\) make \(a = \dfrac{1}{e}.\) This may seem like an odd requirement, but in order to make \(f(x)\) well-defined for both positive and negative values of \(x\) I think that this is necessary.

So with this in mind, we have that \(f(-1) = \displaystyle\int_{-\frac{1}{e}}^{-1} \dfrac{1}{t} dt = \ln |-1| - \ln |-\dfrac{1}{e}| = 1,\) and

\(f(1) = \displaystyle\int_{\frac{1}{e}}^{1} \dfrac{1}{t} dt = \ln(1) - \ln(\dfrac{1}{e}) = 1.\)

So with this interpretation we, in essence, find that \(C = C'\), in agreement with your observation, but I had to rephrase the question so that \(f(x)\) was well-defined for both positive and negative reals in the first place.

I may have confused you even further with this analysis, but feel free to ask any questions you might have. :) – Brian Charlesworth · 1 year, 11 months ago

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Sir @Brian Charlesworth and Sir @Sandeep Bhardwaj please help. – Abhijeet Verma · 1 year, 11 months ago

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