# Constant of integration

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.

Note by Abhijeet Verma
3 years, 2 months ago

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I 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. :)

- 3 years, 2 months ago

- 3 years, 2 months ago

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