How to integrate $frac{cos(x)}{x}$ using substitution












1














Trying to integrate $$int frac{cos(x)}{x} dx = int frac{1}{x}sin'(x) dx$$ by substituting $sin(x)$, but it either becomes more complicated or I end up with a $frac{1}{x}$ still in the integral.






integration substitution







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  • 2




    Not "possible"
    – A.Γ.
    Jan 4 at 13:58


















1














Trying to integrate $$int frac{cos(x)}{x} dx = int frac{1}{x}sin'(x) dx$$ by substituting $sin(x)$, but it either becomes more complicated or I end up with a $frac{1}{x}$ still in the integral.






integration substitution







share|cite|improve this question




















  • 2




    Not "possible"
    – A.Γ.
    Jan 4 at 13:58
















1












1








1


1





Trying to integrate $$int frac{cos(x)}{x} dx = int frac{1}{x}sin'(x) dx$$ by substituting $sin(x)$, but it either becomes more complicated or I end up with a $frac{1}{x}$ still in the integral.






integration substitution







share|cite|improve this question















Trying to integrate $$int frac{cos(x)}{x} dx = int frac{1}{x}sin'(x) dx$$ by substituting $sin(x)$, but it either becomes more complicated or I end up with a $frac{1}{x}$ still in the integral.






integration substitution




integration substitution






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 14:01









A.Γ.

22.6k32656




22.6k32656










asked Jan 4 at 13:56









Conny DagoConny Dago

255




255








  • 2




    Not "possible"
    – A.Γ.
    Jan 4 at 13:58
















  • 2




    Not "possible"
    – A.Γ.
    Jan 4 at 13:58










2




2




Not "possible"
– A.Γ.
Jan 4 at 13:58






Not "possible"
– A.Γ.
Jan 4 at 13:58












2 Answers
2






active

oldest

votes


















3














It is not possible to find an antiderivative of $frac{cos x}{x}$ in term of "elementary functions".



This is a consequence of Liouville's theorem. See link to article for details.






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  • Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
    – Conny Dago
    Jan 4 at 14:16












  • Yes that is correct.
    – mathcounterexamples.net
    Jan 4 at 14:17










  • You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
    – edmz
    Jan 4 at 14:46










  • @edmz You're right! I corrected my answer.
    – mathcounterexamples.net
    Jan 4 at 14:53



















3














As noted the indefinite integral
$$
int frac{cos x}{x};dx
$$
is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $lim_{x to +infty} mathrm{Ci}(x) = 0$. So we may define
$$
mathrm{Ci}(x) = -int_x^inftyfrac{cos t}{t};dt
$$
In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    It is not possible to find an antiderivative of $frac{cos x}{x}$ in term of "elementary functions".



    This is a consequence of Liouville's theorem. See link to article for details.






    share|cite|improve this answer























    • Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
      – Conny Dago
      Jan 4 at 14:16












    • Yes that is correct.
      – mathcounterexamples.net
      Jan 4 at 14:17










    • You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
      – edmz
      Jan 4 at 14:46










    • @edmz You're right! I corrected my answer.
      – mathcounterexamples.net
      Jan 4 at 14:53
















    3














    It is not possible to find an antiderivative of $frac{cos x}{x}$ in term of "elementary functions".



    This is a consequence of Liouville's theorem. See link to article for details.






    share|cite|improve this answer























    • Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
      – Conny Dago
      Jan 4 at 14:16












    • Yes that is correct.
      – mathcounterexamples.net
      Jan 4 at 14:17










    • You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
      – edmz
      Jan 4 at 14:46










    • @edmz You're right! I corrected my answer.
      – mathcounterexamples.net
      Jan 4 at 14:53














    3












    3








    3






    It is not possible to find an antiderivative of $frac{cos x}{x}$ in term of "elementary functions".



    This is a consequence of Liouville's theorem. See link to article for details.






    share|cite|improve this answer














    It is not possible to find an antiderivative of $frac{cos x}{x}$ in term of "elementary functions".



    This is a consequence of Liouville's theorem. See link to article for details.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 4 at 20:44

























    answered Jan 4 at 14:10









    mathcounterexamples.netmathcounterexamples.net

    25.3k21953




    25.3k21953












    • Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
      – Conny Dago
      Jan 4 at 14:16












    • Yes that is correct.
      – mathcounterexamples.net
      Jan 4 at 14:17










    • You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
      – edmz
      Jan 4 at 14:46










    • @edmz You're right! I corrected my answer.
      – mathcounterexamples.net
      Jan 4 at 14:53


















    • Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
      – Conny Dago
      Jan 4 at 14:16












    • Yes that is correct.
      – mathcounterexamples.net
      Jan 4 at 14:17










    • You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
      – edmz
      Jan 4 at 14:46










    • @edmz You're right! I corrected my answer.
      – mathcounterexamples.net
      Jan 4 at 14:53
















    Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
    – Conny Dago
    Jan 4 at 14:16






    Does that mean that integrating $$int frac{1-cos(x)}{x}dx$$ with substitution is also impossible? Because I split it up and thought it would be easier to integrate $frac{1}{x}$ and $frac{cos(x)}{x}$ separately...
    – Conny Dago
    Jan 4 at 14:16














    Yes that is correct.
    – mathcounterexamples.net
    Jan 4 at 14:17




    Yes that is correct.
    – mathcounterexamples.net
    Jan 4 at 14:17












    You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
    – edmz
    Jan 4 at 14:46




    You might want to refer to them as elementary functions, because simple functions often refer to other categories (e.g. in measure theory).
    – edmz
    Jan 4 at 14:46












    @edmz You're right! I corrected my answer.
    – mathcounterexamples.net
    Jan 4 at 14:53




    @edmz You're right! I corrected my answer.
    – mathcounterexamples.net
    Jan 4 at 14:53











    3














    As noted the indefinite integral
    $$
    int frac{cos x}{x};dx
    $$
    is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $lim_{x to +infty} mathrm{Ci}(x) = 0$. So we may define
    $$
    mathrm{Ci}(x) = -int_x^inftyfrac{cos t}{t};dt
    $$
    In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.






    share|cite|improve this answer


























      3














      As noted the indefinite integral
      $$
      int frac{cos x}{x};dx
      $$
      is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $lim_{x to +infty} mathrm{Ci}(x) = 0$. So we may define
      $$
      mathrm{Ci}(x) = -int_x^inftyfrac{cos t}{t};dt
      $$
      In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.






      share|cite|improve this answer
























        3












        3








        3






        As noted the indefinite integral
        $$
        int frac{cos x}{x};dx
        $$
        is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $lim_{x to +infty} mathrm{Ci}(x) = 0$. So we may define
        $$
        mathrm{Ci}(x) = -int_x^inftyfrac{cos t}{t};dt
        $$
        In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.






        share|cite|improve this answer












        As noted the indefinite integral
        $$
        int frac{cos x}{x};dx
        $$
        is not an elementary function. But it is useful enough that it has been given a name, the "cosine integral" function, $mathrm{Ci}(x)$. It is conventional to fix the constant of integration so that $lim_{x to +infty} mathrm{Ci}(x) = 0$. So we may define
        $$
        mathrm{Ci}(x) = -int_x^inftyfrac{cos t}{t};dt
        $$
        In fact, this definition makes sense for $x$ in the complex plane, with a cut along the negative real axis.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 14:40









        GEdgarGEdgar

        62k267168




        62k267168






























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