Integral of Reciprocal Functions: Why The Modulus Sign?












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I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?



I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.



So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.






integration functions







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    1












    $begingroup$


    I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?



    I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.



    So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.






    integration functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?



      I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.



      So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.






      integration functions







      share|cite|improve this question









      $endgroup$




      I've been taught that $intfrac{1}{x}dx=ln|x| + C$ rather than $ln(x) + C$ without the modulus sign and been told that it is incorrect to not have the modulus sign. Why is that?



      I've been investigating the natural logarithms of complex numbers and derived the equation $ln(z)=ln|z|+iarg(z)$ from $z=|z|exp(iθ)$.



      So for negative real numbers this simplifies to $ln(x)=ln|x|+iπ$ so clearly both logarithm functions only differ by a constant so should both be valid functions $intfrac{1}{x}dx$, due to the integration constant being arbitrary.






      integration functions




      integration functions






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      share|cite|improve this question











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      asked Jan 29 at 3:34









      TheTroll73TheTroll73

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          $begingroup$

          This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.



          (All of what follows assumes a real variable. The complex logarithm will not appear.)



          On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.



          In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.






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            $begingroup$

            This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.



            (All of what follows assumes a real variable. The complex logarithm will not appear.)



            On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.



            In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.



              (All of what follows assumes a real variable. The complex logarithm will not appear.)



              On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.



              In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.



                (All of what follows assumes a real variable. The complex logarithm will not appear.)



                On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.



                In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.






                share|cite|improve this answer









                $endgroup$



                This is the sort of philosophical debate that doesn't have a clear right answer. I'll put my view on it out there, for what it's worth.



                (All of what follows assumes a real variable. The complex logarithm will not appear.)



                On $(0,infty)$, $ln x+C$ is an antiderivative for $frac1x$. On $(-infty,0)$, $ln(-x)+C$ is an antiderivative for $frac1x$. We can write these down in one formula as $ln |x|+C$ - but that's deceptive. It creates the false impression that we can integrate across the singularity and get a value for something like $int_{-2}^3 frac1x,dx$ by taking a difference in that antiderivative formula, when in fact that integral diverges.



                In any convergent integral, we won't be crossing that singularity; any integral we can actually evaluate is entirely on one side. However, it's possible that our integral is on the negative side, so we want a general antiderivative that gives us options there. As such, here's my preferred form for the most general antiderivative of $frac1x$: $ln(Ax)$ for nonzero $A$. That's an antiderivative on the positive side if $A>0$, and it's an antiderivative on the negative side if $A<0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 29 at 3:57









                jmerryjmerry

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