Proof of complex integrals involving trigonometic functions












2












$begingroup$


I came across a way of solving the following integrals:
$$I_1=intcosleft[ln xright]dx$$
$$I_2=intsinleft[ln xright]dx$$
firstly:
$$int x^idx=frac{x^{i+1}}{i+1}+C tag{1}$$
and we also know that:
$$int x^idx=int e^{iln(x)}dx=intcosleft[ln xright]+isinleft[ln xright]dxtag{2}$$
Rearranging $(1)$ we get:
$$frac{x^{1+i}}{1+i}=frac{1-i}{2}x^{1+i}=frac{x}{2}left(cosleft[ln xright]+isinleft[ln xright]-icosleft[ln xright]+sinleft[ln xright]right)$$
$$=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+ifrac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}tag{3}$$
and so when we combine $(2)$ and $(3)$ we get:
$$intcosleft[ln xright]dx=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+C$$
$$intsinleft[ln xright]dx=frac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}+C$$
Is this correct? And does anyone have any similar examples of where this method can be used?






integration complex-analysis







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$endgroup$












  • $begingroup$
    You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
    $endgroup$
    – GEdgar
    Jan 15 at 1:39










  • $begingroup$
    Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
    $endgroup$
    – Henry Lee
    Jan 15 at 1:40








  • 1




    $begingroup$
    Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
    $endgroup$
    – GEdgar
    Jan 15 at 1:43
















2












$begingroup$


I came across a way of solving the following integrals:
$$I_1=intcosleft[ln xright]dx$$
$$I_2=intsinleft[ln xright]dx$$
firstly:
$$int x^idx=frac{x^{i+1}}{i+1}+C tag{1}$$
and we also know that:
$$int x^idx=int e^{iln(x)}dx=intcosleft[ln xright]+isinleft[ln xright]dxtag{2}$$
Rearranging $(1)$ we get:
$$frac{x^{1+i}}{1+i}=frac{1-i}{2}x^{1+i}=frac{x}{2}left(cosleft[ln xright]+isinleft[ln xright]-icosleft[ln xright]+sinleft[ln xright]right)$$
$$=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+ifrac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}tag{3}$$
and so when we combine $(2)$ and $(3)$ we get:
$$intcosleft[ln xright]dx=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+C$$
$$intsinleft[ln xright]dx=frac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}+C$$
Is this correct? And does anyone have any similar examples of where this method can be used?






integration complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
    $endgroup$
    – GEdgar
    Jan 15 at 1:39










  • $begingroup$
    Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
    $endgroup$
    – Henry Lee
    Jan 15 at 1:40








  • 1




    $begingroup$
    Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
    $endgroup$
    – GEdgar
    Jan 15 at 1:43














2












2








2





$begingroup$


I came across a way of solving the following integrals:
$$I_1=intcosleft[ln xright]dx$$
$$I_2=intsinleft[ln xright]dx$$
firstly:
$$int x^idx=frac{x^{i+1}}{i+1}+C tag{1}$$
and we also know that:
$$int x^idx=int e^{iln(x)}dx=intcosleft[ln xright]+isinleft[ln xright]dxtag{2}$$
Rearranging $(1)$ we get:
$$frac{x^{1+i}}{1+i}=frac{1-i}{2}x^{1+i}=frac{x}{2}left(cosleft[ln xright]+isinleft[ln xright]-icosleft[ln xright]+sinleft[ln xright]right)$$
$$=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+ifrac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}tag{3}$$
and so when we combine $(2)$ and $(3)$ we get:
$$intcosleft[ln xright]dx=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+C$$
$$intsinleft[ln xright]dx=frac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}+C$$
Is this correct? And does anyone have any similar examples of where this method can be used?






integration complex-analysis







share|cite|improve this question











$endgroup$




I came across a way of solving the following integrals:
$$I_1=intcosleft[ln xright]dx$$
$$I_2=intsinleft[ln xright]dx$$
firstly:
$$int x^idx=frac{x^{i+1}}{i+1}+C tag{1}$$
and we also know that:
$$int x^idx=int e^{iln(x)}dx=intcosleft[ln xright]+isinleft[ln xright]dxtag{2}$$
Rearranging $(1)$ we get:
$$frac{x^{1+i}}{1+i}=frac{1-i}{2}x^{1+i}=frac{x}{2}left(cosleft[ln xright]+isinleft[ln xright]-icosleft[ln xright]+sinleft[ln xright]right)$$
$$=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+ifrac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}tag{3}$$
and so when we combine $(2)$ and $(3)$ we get:
$$intcosleft[ln xright]dx=frac{xleft(cosleft[ln xright]+sinleft[ln xright]right)}{2}+C$$
$$intsinleft[ln xright]dx=frac{xleft(sinleft[ln xright]-cosleft[ln xright]right)}{2}+C$$
Is this correct? And does anyone have any similar examples of where this method can be used?






integration complex-analysis




integration complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 1:39







Henry Lee

















asked Jan 15 at 1:37









Henry LeeHenry Lee

2,181319




2,181319












  • $begingroup$
    You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
    $endgroup$
    – GEdgar
    Jan 15 at 1:39










  • $begingroup$
    Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
    $endgroup$
    – Henry Lee
    Jan 15 at 1:40








  • 1




    $begingroup$
    Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
    $endgroup$
    – GEdgar
    Jan 15 at 1:43


















  • $begingroup$
    You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
    $endgroup$
    – GEdgar
    Jan 15 at 1:39










  • $begingroup$
    Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
    $endgroup$
    – Henry Lee
    Jan 15 at 1:40








  • 1




    $begingroup$
    Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
    $endgroup$
    – GEdgar
    Jan 15 at 1:43
















$begingroup$
You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
$endgroup$
– GEdgar
Jan 15 at 1:39




$begingroup$
You can check yourself whether attempted anti-derivatives are correct. Differentiate the right-hand-side, and see if the result is the integrand.
$endgroup$
– GEdgar
Jan 15 at 1:39












$begingroup$
Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
$endgroup$
– Henry Lee
Jan 15 at 1:40






$begingroup$
Yes when differentiated it gives the desired result, but are there any limitation that I have overlooked, other than the function not being continuous for real
$endgroup$
– Henry Lee
Jan 15 at 1:40






1




1




$begingroup$
Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
$endgroup$
– GEdgar
Jan 15 at 1:43




$begingroup$
Your integrand must have $x>0$. Then $x^i$ is defined and has the continuous principal value you wrote.
$endgroup$
– GEdgar
Jan 15 at 1:43










1 Answer
1






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1












$begingroup$

Recall
$$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
Hence
$$begin{align}
sin log x=&frac{e^{ilog x}-e^{-ilog x}}{2i}\
=&frac{x^{i}-x^{-i}}{2i}\
end{align}$$
Similarly:
$$cos log x=frac{x^i+x^{-i}}2$$
each of which can be easily integrated giving the desired results.






share|cite|improve this answer









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    $begingroup$

    Recall
    $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
    Hence
    $$begin{align}
    sin log x=&frac{e^{ilog x}-e^{-ilog x}}{2i}\
    =&frac{x^{i}-x^{-i}}{2i}\
    end{align}$$
    Similarly:
    $$cos log x=frac{x^i+x^{-i}}2$$
    each of which can be easily integrated giving the desired results.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Recall
      $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
      Hence
      $$begin{align}
      sin log x=&frac{e^{ilog x}-e^{-ilog x}}{2i}\
      =&frac{x^{i}-x^{-i}}{2i}\
      end{align}$$
      Similarly:
      $$cos log x=frac{x^i+x^{-i}}2$$
      each of which can be easily integrated giving the desired results.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Recall
        $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
        Hence
        $$begin{align}
        sin log x=&frac{e^{ilog x}-e^{-ilog x}}{2i}\
        =&frac{x^{i}-x^{-i}}{2i}\
        end{align}$$
        Similarly:
        $$cos log x=frac{x^i+x^{-i}}2$$
        each of which can be easily integrated giving the desired results.






        share|cite|improve this answer









        $endgroup$



        Recall
        $$sin x=frac{e^{ix}-e^{-ix}}{2i}$$
        Hence
        $$begin{align}
        sin log x=&frac{e^{ilog x}-e^{-ilog x}}{2i}\
        =&frac{x^{i}-x^{-i}}{2i}\
        end{align}$$
        Similarly:
        $$cos log x=frac{x^i+x^{-i}}2$$
        each of which can be easily integrated giving the desired results.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 3:15









        clathratusclathratus

        4,9641438




        4,9641438






























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