Prove that $intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$ converges












3












$begingroup$



Prove the convergence of



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$




First I thought the integral does not converge because



$$intlimits_1^{infty} -frac{1}{x} ,mathrm{d}x le intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$



But in this case



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x le intlimits_1^{infty} frac{1}{x^2} , mathrm{d}x$$



it converges concerning the majorant criterion. What's the right way?






calculus real-analysis convergence improper-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Dirichlet%27s_test
    $endgroup$
    – Jack D'Aurizio
    Jul 15 '16 at 12:26






  • 1




    $begingroup$
    If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
    $endgroup$
    – StackTD
    Jul 15 '16 at 12:28










  • $begingroup$
    $f,g:[0,infty]to mathbb{R}$, is this the problem?
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:32












  • $begingroup$
    @jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
    $endgroup$
    – 5xum
    Jul 15 '16 at 12:33












  • $begingroup$
    @5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:35
















3












$begingroup$



Prove the convergence of



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$




First I thought the integral does not converge because



$$intlimits_1^{infty} -frac{1}{x} ,mathrm{d}x le intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$



But in this case



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x le intlimits_1^{infty} frac{1}{x^2} , mathrm{d}x$$



it converges concerning the majorant criterion. What's the right way?






calculus real-analysis convergence improper-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Dirichlet%27s_test
    $endgroup$
    – Jack D'Aurizio
    Jul 15 '16 at 12:26






  • 1




    $begingroup$
    If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
    $endgroup$
    – StackTD
    Jul 15 '16 at 12:28










  • $begingroup$
    $f,g:[0,infty]to mathbb{R}$, is this the problem?
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:32












  • $begingroup$
    @jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
    $endgroup$
    – 5xum
    Jul 15 '16 at 12:33












  • $begingroup$
    @5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:35














3












3








3


2



$begingroup$



Prove the convergence of



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$




First I thought the integral does not converge because



$$intlimits_1^{infty} -frac{1}{x} ,mathrm{d}x le intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$



But in this case



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x le intlimits_1^{infty} frac{1}{x^2} , mathrm{d}x$$



it converges concerning the majorant criterion. What's the right way?






calculus real-analysis convergence improper-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$





Prove the convergence of



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$




First I thought the integral does not converge because



$$intlimits_1^{infty} -frac{1}{x} ,mathrm{d}x le intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x$$



But in this case



$$intlimits_1^{infty} frac{cos(x)}{x} , mathrm{d}x le intlimits_1^{infty} frac{1}{x^2} , mathrm{d}x$$



it converges concerning the majorant criterion. What's the right way?






calculus real-analysis convergence improper-integrals trigonometric-integrals




calculus real-analysis convergence improper-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 15 '16 at 12:36









Rodrigo de Azevedo

13k41958




13k41958










asked Jul 15 '16 at 12:25









jacmeirdjacmeird

503315




503315












  • $begingroup$
    en.wikipedia.org/wiki/Dirichlet%27s_test
    $endgroup$
    – Jack D'Aurizio
    Jul 15 '16 at 12:26






  • 1




    $begingroup$
    If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
    $endgroup$
    – StackTD
    Jul 15 '16 at 12:28










  • $begingroup$
    $f,g:[0,infty]to mathbb{R}$, is this the problem?
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:32












  • $begingroup$
    @jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
    $endgroup$
    – 5xum
    Jul 15 '16 at 12:33












  • $begingroup$
    @5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:35


















  • $begingroup$
    en.wikipedia.org/wiki/Dirichlet%27s_test
    $endgroup$
    – Jack D'Aurizio
    Jul 15 '16 at 12:26






  • 1




    $begingroup$
    If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
    $endgroup$
    – StackTD
    Jul 15 '16 at 12:28










  • $begingroup$
    $f,g:[0,infty]to mathbb{R}$, is this the problem?
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:32












  • $begingroup$
    @jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
    $endgroup$
    – 5xum
    Jul 15 '16 at 12:33












  • $begingroup$
    @5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
    $endgroup$
    – jacmeird
    Jul 15 '16 at 12:35
















$begingroup$
en.wikipedia.org/wiki/Dirichlet%27s_test
$endgroup$
– Jack D'Aurizio
Jul 15 '16 at 12:26




$begingroup$
en.wikipedia.org/wiki/Dirichlet%27s_test
$endgroup$
– Jack D'Aurizio
Jul 15 '16 at 12:26




1




1




$begingroup$
If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
$endgroup$
– StackTD
Jul 15 '16 at 12:28




$begingroup$
If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions).
$endgroup$
– StackTD
Jul 15 '16 at 12:28












$begingroup$
$f,g:[0,infty]to mathbb{R}$, is this the problem?
$endgroup$
– jacmeird
Jul 15 '16 at 12:32






$begingroup$
$f,g:[0,infty]to mathbb{R}$, is this the problem?
$endgroup$
– jacmeird
Jul 15 '16 at 12:32














$begingroup$
@jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
$endgroup$
– 5xum
Jul 15 '16 at 12:33






$begingroup$
@jacmeird The problem is that $frac{cos x}{x}leq frac{1}{x^2}$ is not true! Take any $x>1$ such that $cos x = 1$ and you will see.
$endgroup$
– 5xum
Jul 15 '16 at 12:33














$begingroup$
@5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
$endgroup$
– jacmeird
Jul 15 '16 at 12:35




$begingroup$
@5xum Ouch, of course you're right. If I take $frac{cos(x)}{x} le frac{1}{x}$, there is the same problem.
$endgroup$
– jacmeird
Jul 15 '16 at 12:35










3 Answers
3






active

oldest

votes


















5












$begingroup$

You might want to use integration by parts, obtaining for $Mge1$,
$$
int_{1}^M frac{cos x}{x}: dx=left[frac{sin x}{ x}right]_1^M+ int_1^M frac{sin x}{x^2}: dx
$$ letting $M to infty$ gives
$$
int_{1}^infty frac{cos x}{x} :dx=lim_{M to infty}int_1^M frac{cos x}{x} :dx= -sin 1+int_1^infty frac{sin x}{ x^2}: dx
$$then one may conclude by the absolute convergence of the latter integral:
$$
left|int_1^infty frac{sin x}{ x^2}: dxright|<int_1^infty frac{|sin x|}{ x^2}: dx<int_1^infty frac{1}{x^2}: dx<infty.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
    $endgroup$
    – Mc Cheng
    Jul 15 '16 at 12:43










  • $begingroup$
    @McCheng Edited. Thank you very much!
    $endgroup$
    – Olivier Oloa
    Jul 15 '16 at 12:44






  • 1




    $begingroup$
    You should leave off those absolute values in your last line.
    $endgroup$
    – zhw.
    Jul 15 '16 at 15:56










  • $begingroup$
    @zhw. You are right, done.
    $endgroup$
    – Olivier Oloa
    Jan 9 at 8:24










  • $begingroup$
    I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
    $endgroup$
    – Ixion
    Jan 9 at 10:48





















0












$begingroup$

Hint: Awful and tricky but as the problem is near $+infty$, write $$int limits_{frac{pi}{2}}^{Npi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x= sum limits_{k=1}^{N}left(intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}xright)$$ and use the criterion for alternating series with the sequence $$a_k=intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $$
    begin{align}
    left|,int_1^inftyfrac{cos(x)}{x},mathrm{d}x,right|
    &leleft|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}frac{cos(x)}{x},mathrm{d}x,right|\
    &=left|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}cos(x)left(frac1{x}-frac1{2kpi}right),mathrm{d}x,right|\
    &lelog(pi)+sum_{k=1}^infty2pileft(frac1{(2k-1)pi}-frac1{2kpi}right)\[6pt]
    &=log(4pi)
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      You might want to use integration by parts, obtaining for $Mge1$,
      $$
      int_{1}^M frac{cos x}{x}: dx=left[frac{sin x}{ x}right]_1^M+ int_1^M frac{sin x}{x^2}: dx
      $$       letting $M to infty$ gives
      $$
      int_{1}^infty frac{cos x}{x} :dx=lim_{M to infty}int_1^M frac{cos x}{x} :dx= -sin 1+int_1^infty frac{sin x}{ x^2}: dx
      $$      then one may conclude by the absolute convergence of the latter integral:
      $$
      left|int_1^infty frac{sin x}{ x^2}: dxright|<int_1^infty frac{|sin x|}{ x^2}: dx<int_1^infty frac{1}{x^2}: dx<infty.
      $$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
        $endgroup$
        – Mc Cheng
        Jul 15 '16 at 12:43










      • $begingroup$
        @McCheng Edited. Thank you very much!
        $endgroup$
        – Olivier Oloa
        Jul 15 '16 at 12:44






      • 1




        $begingroup$
        You should leave off those absolute values in your last line.
        $endgroup$
        – zhw.
        Jul 15 '16 at 15:56










      • $begingroup$
        @zhw. You are right, done.
        $endgroup$
        – Olivier Oloa
        Jan 9 at 8:24










      • $begingroup$
        I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
        $endgroup$
        – Ixion
        Jan 9 at 10:48


















      5












      $begingroup$

      You might want to use integration by parts, obtaining for $Mge1$,
      $$
      int_{1}^M frac{cos x}{x}: dx=left[frac{sin x}{ x}right]_1^M+ int_1^M frac{sin x}{x^2}: dx
      $$       letting $M to infty$ gives
      $$
      int_{1}^infty frac{cos x}{x} :dx=lim_{M to infty}int_1^M frac{cos x}{x} :dx= -sin 1+int_1^infty frac{sin x}{ x^2}: dx
      $$      then one may conclude by the absolute convergence of the latter integral:
      $$
      left|int_1^infty frac{sin x}{ x^2}: dxright|<int_1^infty frac{|sin x|}{ x^2}: dx<int_1^infty frac{1}{x^2}: dx<infty.
      $$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
        $endgroup$
        – Mc Cheng
        Jul 15 '16 at 12:43










      • $begingroup$
        @McCheng Edited. Thank you very much!
        $endgroup$
        – Olivier Oloa
        Jul 15 '16 at 12:44






      • 1




        $begingroup$
        You should leave off those absolute values in your last line.
        $endgroup$
        – zhw.
        Jul 15 '16 at 15:56










      • $begingroup$
        @zhw. You are right, done.
        $endgroup$
        – Olivier Oloa
        Jan 9 at 8:24










      • $begingroup$
        I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
        $endgroup$
        – Ixion
        Jan 9 at 10:48
















      5












      5








      5





      $begingroup$

      You might want to use integration by parts, obtaining for $Mge1$,
      $$
      int_{1}^M frac{cos x}{x}: dx=left[frac{sin x}{ x}right]_1^M+ int_1^M frac{sin x}{x^2}: dx
      $$       letting $M to infty$ gives
      $$
      int_{1}^infty frac{cos x}{x} :dx=lim_{M to infty}int_1^M frac{cos x}{x} :dx= -sin 1+int_1^infty frac{sin x}{ x^2}: dx
      $$      then one may conclude by the absolute convergence of the latter integral:
      $$
      left|int_1^infty frac{sin x}{ x^2}: dxright|<int_1^infty frac{|sin x|}{ x^2}: dx<int_1^infty frac{1}{x^2}: dx<infty.
      $$






      share|cite|improve this answer











      $endgroup$



      You might want to use integration by parts, obtaining for $Mge1$,
      $$
      int_{1}^M frac{cos x}{x}: dx=left[frac{sin x}{ x}right]_1^M+ int_1^M frac{sin x}{x^2}: dx
      $$       letting $M to infty$ gives
      $$
      int_{1}^infty frac{cos x}{x} :dx=lim_{M to infty}int_1^M frac{cos x}{x} :dx= -sin 1+int_1^infty frac{sin x}{ x^2}: dx
      $$      then one may conclude by the absolute convergence of the latter integral:
      $$
      left|int_1^infty frac{sin x}{ x^2}: dxright|<int_1^infty frac{|sin x|}{ x^2}: dx<int_1^infty frac{1}{x^2}: dx<infty.
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 11 at 4:38

























      answered Jul 15 '16 at 12:36









      Olivier OloaOlivier Oloa

      108k17177294




      108k17177294












      • $begingroup$
        Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
        $endgroup$
        – Mc Cheng
        Jul 15 '16 at 12:43










      • $begingroup$
        @McCheng Edited. Thank you very much!
        $endgroup$
        – Olivier Oloa
        Jul 15 '16 at 12:44






      • 1




        $begingroup$
        You should leave off those absolute values in your last line.
        $endgroup$
        – zhw.
        Jul 15 '16 at 15:56










      • $begingroup$
        @zhw. You are right, done.
        $endgroup$
        – Olivier Oloa
        Jan 9 at 8:24










      • $begingroup$
        I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
        $endgroup$
        – Ixion
        Jan 9 at 10:48




















      • $begingroup$
        Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
        $endgroup$
        – Mc Cheng
        Jul 15 '16 at 12:43










      • $begingroup$
        @McCheng Edited. Thank you very much!
        $endgroup$
        – Olivier Oloa
        Jul 15 '16 at 12:44






      • 1




        $begingroup$
        You should leave off those absolute values in your last line.
        $endgroup$
        – zhw.
        Jul 15 '16 at 15:56










      • $begingroup$
        @zhw. You are right, done.
        $endgroup$
        – Olivier Oloa
        Jan 9 at 8:24










      • $begingroup$
        I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
        $endgroup$
        – Ixion
        Jan 9 at 10:48


















      $begingroup$
      Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
      $endgroup$
      – Mc Cheng
      Jul 15 '16 at 12:43




      $begingroup$
      Is it $left|lim_{M to infty}int_{1}^M frac{cos x}{x} dxright|$ ?
      $endgroup$
      – Mc Cheng
      Jul 15 '16 at 12:43












      $begingroup$
      @McCheng Edited. Thank you very much!
      $endgroup$
      – Olivier Oloa
      Jul 15 '16 at 12:44




      $begingroup$
      @McCheng Edited. Thank you very much!
      $endgroup$
      – Olivier Oloa
      Jul 15 '16 at 12:44




      1




      1




      $begingroup$
      You should leave off those absolute values in your last line.
      $endgroup$
      – zhw.
      Jul 15 '16 at 15:56




      $begingroup$
      You should leave off those absolute values in your last line.
      $endgroup$
      – zhw.
      Jul 15 '16 at 15:56












      $begingroup$
      @zhw. You are right, done.
      $endgroup$
      – Olivier Oloa
      Jan 9 at 8:24




      $begingroup$
      @zhw. You are right, done.
      $endgroup$
      – Olivier Oloa
      Jan 9 at 8:24












      $begingroup$
      I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
      $endgroup$
      – Ixion
      Jan 9 at 10:48






      $begingroup$
      I think you lost a minus sign in front of $frac{sin 1}{1}$, right?
      $endgroup$
      – Ixion
      Jan 9 at 10:48













      0












      $begingroup$

      Hint: Awful and tricky but as the problem is near $+infty$, write $$int limits_{frac{pi}{2}}^{Npi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x= sum limits_{k=1}^{N}left(intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}xright)$$ and use the criterion for alternating series with the sequence $$a_k=intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x$$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Hint: Awful and tricky but as the problem is near $+infty$, write $$int limits_{frac{pi}{2}}^{Npi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x= sum limits_{k=1}^{N}left(intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}xright)$$ and use the criterion for alternating series with the sequence $$a_k=intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x$$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint: Awful and tricky but as the problem is near $+infty$, write $$int limits_{frac{pi}{2}}^{Npi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x= sum limits_{k=1}^{N}left(intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}xright)$$ and use the criterion for alternating series with the sequence $$a_k=intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x$$






          share|cite|improve this answer











          $endgroup$



          Hint: Awful and tricky but as the problem is near $+infty$, write $$int limits_{frac{pi}{2}}^{Npi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x= sum limits_{k=1}^{N}left(intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}xright)$$ and use the criterion for alternating series with the sequence $$a_k=intlimits_{kpi-frac{pi}{2}}^{kpi+frac{pi}{2}}frac{cos(x)}{x}mathrm{d}x$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 9 at 9:01









          Lorenzo B.

          1,8402520




          1,8402520










          answered Jan 6 at 21:23









          MamanMaman

          1,194722




          1,194722























              0












              $begingroup$

              $$
              begin{align}
              left|,int_1^inftyfrac{cos(x)}{x},mathrm{d}x,right|
              &leleft|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}frac{cos(x)}{x},mathrm{d}x,right|\
              &=left|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}cos(x)left(frac1{x}-frac1{2kpi}right),mathrm{d}x,right|\
              &lelog(pi)+sum_{k=1}^infty2pileft(frac1{(2k-1)pi}-frac1{2kpi}right)\[6pt]
              &=log(4pi)
              end{align}
              $$






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              $endgroup$


















                0












                $begingroup$

                $$
                begin{align}
                left|,int_1^inftyfrac{cos(x)}{x},mathrm{d}x,right|
                &leleft|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}frac{cos(x)}{x},mathrm{d}x,right|\
                &=left|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}cos(x)left(frac1{x}-frac1{2kpi}right),mathrm{d}x,right|\
                &lelog(pi)+sum_{k=1}^infty2pileft(frac1{(2k-1)pi}-frac1{2kpi}right)\[6pt]
                &=log(4pi)
                end{align}
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$
                  begin{align}
                  left|,int_1^inftyfrac{cos(x)}{x},mathrm{d}x,right|
                  &leleft|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}frac{cos(x)}{x},mathrm{d}x,right|\
                  &=left|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}cos(x)left(frac1{x}-frac1{2kpi}right),mathrm{d}x,right|\
                  &lelog(pi)+sum_{k=1}^infty2pileft(frac1{(2k-1)pi}-frac1{2kpi}right)\[6pt]
                  &=log(4pi)
                  end{align}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  $$
                  begin{align}
                  left|,int_1^inftyfrac{cos(x)}{x},mathrm{d}x,right|
                  &leleft|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}frac{cos(x)}{x},mathrm{d}x,right|\
                  &=left|,int_1^pifrac{cos(x)}{x},mathrm{d}x,right|+sum_{k=1}^inftyleft|,int_{(2k-1)pi}^{(2k+1)pi}cos(x)left(frac1{x}-frac1{2kpi}right),mathrm{d}x,right|\
                  &lelog(pi)+sum_{k=1}^infty2pileft(frac1{(2k-1)pi}-frac1{2kpi}right)\[6pt]
                  &=log(4pi)
                  end{align}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 10:59









                  robjohn♦robjohn

                  268k27308632




                  268k27308632






























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