Evaluating $int_1^infty frac{1}{x(x^2+1)} dx$












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$begingroup$


$$I=int_1^infty frac{1}{x(x^2+1)} dx$$



I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.



$$frac{1}{x(x^2+1)} = frac{1}{2} left(frac{1}{x}-frac{1}{x+i} -frac{1}{x-i}right)$$

So
$$I = frac{1}{2}int_1^inftyfrac{1}{x}-frac{1}{x+i} -frac{1}{x-i} dx \ = frac{1}{2}left[log |x| - log(|x+i|) - log(|x-i|)right]_1^infty
$$
which evaluates to be $infty$.






integration proof-verification improper-integrals







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  • 4




    $begingroup$
    Try 1/x - x/(x^2+1)
    $endgroup$
    – jnyan
    Nov 23 '17 at 6:03










  • $begingroup$
    The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
    $endgroup$
    – Claude Leibovici
    Nov 23 '17 at 6:07










  • $begingroup$
    @ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
    $endgroup$
    – Natash1
    Nov 23 '17 at 6:11






  • 2




    $begingroup$
    You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
    $endgroup$
    – Sangchul Lee
    Nov 23 '17 at 6:15








  • 1




    $begingroup$
    Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
    $endgroup$
    – Nosrati
    Dec 11 '18 at 11:06
















1












$begingroup$


$$I=int_1^infty frac{1}{x(x^2+1)} dx$$



I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.



$$frac{1}{x(x^2+1)} = frac{1}{2} left(frac{1}{x}-frac{1}{x+i} -frac{1}{x-i}right)$$

So
$$I = frac{1}{2}int_1^inftyfrac{1}{x}-frac{1}{x+i} -frac{1}{x-i} dx \ = frac{1}{2}left[log |x| - log(|x+i|) - log(|x-i|)right]_1^infty
$$
which evaluates to be $infty$.






integration proof-verification improper-integrals







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  • 4




    $begingroup$
    Try 1/x - x/(x^2+1)
    $endgroup$
    – jnyan
    Nov 23 '17 at 6:03










  • $begingroup$
    The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
    $endgroup$
    – Claude Leibovici
    Nov 23 '17 at 6:07










  • $begingroup$
    @ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
    $endgroup$
    – Natash1
    Nov 23 '17 at 6:11






  • 2




    $begingroup$
    You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
    $endgroup$
    – Sangchul Lee
    Nov 23 '17 at 6:15








  • 1




    $begingroup$
    Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
    $endgroup$
    – Nosrati
    Dec 11 '18 at 11:06














1












1








1


1



$begingroup$


$$I=int_1^infty frac{1}{x(x^2+1)} dx$$



I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.



$$frac{1}{x(x^2+1)} = frac{1}{2} left(frac{1}{x}-frac{1}{x+i} -frac{1}{x-i}right)$$

So
$$I = frac{1}{2}int_1^inftyfrac{1}{x}-frac{1}{x+i} -frac{1}{x-i} dx \ = frac{1}{2}left[log |x| - log(|x+i|) - log(|x-i|)right]_1^infty
$$
which evaluates to be $infty$.






integration proof-verification improper-integrals







share|cite|improve this question











$endgroup$




$$I=int_1^infty frac{1}{x(x^2+1)} dx$$



I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.



$$frac{1}{x(x^2+1)} = frac{1}{2} left(frac{1}{x}-frac{1}{x+i} -frac{1}{x-i}right)$$

So
$$I = frac{1}{2}int_1^inftyfrac{1}{x}-frac{1}{x+i} -frac{1}{x-i} dx \ = frac{1}{2}left[log |x| - log(|x+i|) - log(|x-i|)right]_1^infty
$$
which evaluates to be $infty$.






integration proof-verification improper-integrals




integration proof-verification improper-integrals






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share|cite|improve this question













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edited Dec 11 '18 at 12:08









Martin Sleziak

44.8k9118272




44.8k9118272










asked Nov 23 '17 at 5:59









Natash1Natash1

619213




619213








  • 4




    $begingroup$
    Try 1/x - x/(x^2+1)
    $endgroup$
    – jnyan
    Nov 23 '17 at 6:03










  • $begingroup$
    The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
    $endgroup$
    – Claude Leibovici
    Nov 23 '17 at 6:07










  • $begingroup$
    @ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
    $endgroup$
    – Natash1
    Nov 23 '17 at 6:11






  • 2




    $begingroup$
    You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
    $endgroup$
    – Sangchul Lee
    Nov 23 '17 at 6:15








  • 1




    $begingroup$
    Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
    $endgroup$
    – Nosrati
    Dec 11 '18 at 11:06














  • 4




    $begingroup$
    Try 1/x - x/(x^2+1)
    $endgroup$
    – jnyan
    Nov 23 '17 at 6:03










  • $begingroup$
    The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
    $endgroup$
    – Claude Leibovici
    Nov 23 '17 at 6:07










  • $begingroup$
    @ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
    $endgroup$
    – Natash1
    Nov 23 '17 at 6:11






  • 2




    $begingroup$
    You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
    $endgroup$
    – Sangchul Lee
    Nov 23 '17 at 6:15








  • 1




    $begingroup$
    Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
    $endgroup$
    – Nosrati
    Dec 11 '18 at 11:06








4




4




$begingroup$
Try 1/x - x/(x^2+1)
$endgroup$
– jnyan
Nov 23 '17 at 6:03




$begingroup$
Try 1/x - x/(x^2+1)
$endgroup$
– jnyan
Nov 23 '17 at 6:03












$begingroup$
The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
$endgroup$
– Claude Leibovici
Nov 23 '17 at 6:07




$begingroup$
The factor $frac 12$ does not apply for $frac 1x$. Otherwise, you are on the right track if you simplify the complex terms.
$endgroup$
– Claude Leibovici
Nov 23 '17 at 6:07












$begingroup$
@ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
$endgroup$
– Natash1
Nov 23 '17 at 6:11




$begingroup$
@ClaudeLeibovici How about if I don't simplify the complex terms (just substituting the limits in)? Wouldn't I get something like $-infty - infty$ (a sum of infinities)?
$endgroup$
– Natash1
Nov 23 '17 at 6:11




2




2




$begingroup$
You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
$endgroup$
– Sangchul Lee
Nov 23 '17 at 6:15






$begingroup$
You end up with $$ I = int_{1}^{infty} left( frac{1}{x} - frac{x}{1+x^2} right) , dx = int_{1}^{infty} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ And notice that this is the limit $$ I = lim_{Rtoinfty} int_{1}^{R} left( frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} right) , dx. $$ This does help in evaluating $I$ since you now have a direct control over the indeterminate of the form $infty - infty$.
$endgroup$
– Sangchul Lee
Nov 23 '17 at 6:15






1




1




$begingroup$
Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:06




$begingroup$
Possible duplicate of Evaluation of $int_1^infty frac{1}{x(x^2+1)} dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:06










5 Answers
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$begingroup$

Why complicate it with complex coefficients?
$$int_1^inftyfrac{1}{x(x^2+1)}dx=int_1^inftyleft(frac{1}{x}-frac{x}{x^2+1}right)dx=left(ln(x)-frac{1}{2}ln(x^2+1)right)biggvert_1^infty\=lnfrac{x}{sqrt{x^2+1}}biggvert_1^infty=lnfrac{1}{sqrt{1+1/x^2}}biggvert_1^infty=ln 1-lnfrac{1}{sqrt{2}}=lnsqrt 2$$






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  • $begingroup$
    While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
    $endgroup$
    – Dylan
    Nov 23 '17 at 7:33



















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$begingroup$

First of all:
$$frac{1}{x(x^2+1)} = frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)$$



Second of all:



You have to evaluate the integral $int_1 ^ Rfrac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $infty -infty$ is undetermined.



$$int_1 ^ Rfrac{1}{x(x^2+1)} = int_1 ^ R left[frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)right]dx$$
$$=[log |x| - frac{1}2{}log(|x+i|) - frac{1}{2}log(|x-i|)]_1^R$$



$$=[log |R| - frac{1}2{}log(|R+i|) - frac{1}{2}log(|R-i|)] + frac{1}2{}log(|1+i|) + frac{1}{2}log(|1-i|)]$$



$$=logleft|frac{R}{(R+i)^{frac{1}{2}}(R-i)^frac{1}{2}}right|+frac{1}{2}log(1+i)(1-i)$$
$$=logleft|frac{R}{(R^2+1)^frac{1}{2}}right|+frac{1}{2}log(2)$$
Now taking the limit yields $frac{1}{2}log(2)$.






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    $begingroup$

    put $x = tan theta$ given integral is
    $$int_{frac{pi}{4}}^{frac{pi}{2}} frac{sec^2 theta}{tan theta sec^2 theta} = frac{d(sin theta)}{sin theta} = ln [|sin theta|]^{frac{pi}{2}}_{frac{pi}{4}} = 0-ln frac{1}{sqrt{2}}=ln sqrt{2}$$






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      2












      $begingroup$

      Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's



      $$ frac{1}{x(1+x^2)} = frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} $$



      Then



      $$ int_1^{infty} f(x) dx = left.left(ln |x| - frac{1}{2}ln |x+i| -
      frac{1}{2} ln |x-i|right)right|_1^{infty} = left. ln left|frac{x}{sqrt{x^2+1}} right|right|_1^{infty} = frac{1}{2}ln 2 $$



      Note that
      $$ lim_{xto infty} frac{x}{sqrt{x^2+1}} = lim_{xtoinfty}frac{1}{sqrt{1+frac{1}{x^2}}} = 1 $$






      share|cite|improve this answer











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      • 1




        $begingroup$
        The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
        $endgroup$
        – Andrei
        Nov 23 '17 at 8:09










      • $begingroup$
        Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
        $endgroup$
        – Dylan
        Nov 23 '17 at 9:08












      • $begingroup$
        No problem. Happens to everyone from time to time
        $endgroup$
        – Andrei
        Nov 23 '17 at 9:20



















      2












      $begingroup$

      By enforcing the substitution $x=frac{1}{z}$ the result is straightforward to find:



      $$ int_{1}^{+infty}frac{dx}{x(x^2+1)}=int_{0}^{1}frac{z,dz}{z^2+1}=left[frac{1}{2}log(1+z^2)right]_{0}^{1}=color{red}{frac{log 2}{2}}.$$






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        5 Answers
        5






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        5 Answers
        5






        active

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        active

        oldest

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        2












        $begingroup$

        Why complicate it with complex coefficients?
        $$int_1^inftyfrac{1}{x(x^2+1)}dx=int_1^inftyleft(frac{1}{x}-frac{x}{x^2+1}right)dx=left(ln(x)-frac{1}{2}ln(x^2+1)right)biggvert_1^infty\=lnfrac{x}{sqrt{x^2+1}}biggvert_1^infty=lnfrac{1}{sqrt{1+1/x^2}}biggvert_1^infty=ln 1-lnfrac{1}{sqrt{2}}=lnsqrt 2$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
          $endgroup$
          – Dylan
          Nov 23 '17 at 7:33
















        2












        $begingroup$

        Why complicate it with complex coefficients?
        $$int_1^inftyfrac{1}{x(x^2+1)}dx=int_1^inftyleft(frac{1}{x}-frac{x}{x^2+1}right)dx=left(ln(x)-frac{1}{2}ln(x^2+1)right)biggvert_1^infty\=lnfrac{x}{sqrt{x^2+1}}biggvert_1^infty=lnfrac{1}{sqrt{1+1/x^2}}biggvert_1^infty=ln 1-lnfrac{1}{sqrt{2}}=lnsqrt 2$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
          $endgroup$
          – Dylan
          Nov 23 '17 at 7:33














        2












        2








        2





        $begingroup$

        Why complicate it with complex coefficients?
        $$int_1^inftyfrac{1}{x(x^2+1)}dx=int_1^inftyleft(frac{1}{x}-frac{x}{x^2+1}right)dx=left(ln(x)-frac{1}{2}ln(x^2+1)right)biggvert_1^infty\=lnfrac{x}{sqrt{x^2+1}}biggvert_1^infty=lnfrac{1}{sqrt{1+1/x^2}}biggvert_1^infty=ln 1-lnfrac{1}{sqrt{2}}=lnsqrt 2$$






        share|cite|improve this answer









        $endgroup$



        Why complicate it with complex coefficients?
        $$int_1^inftyfrac{1}{x(x^2+1)}dx=int_1^inftyleft(frac{1}{x}-frac{x}{x^2+1}right)dx=left(ln(x)-frac{1}{2}ln(x^2+1)right)biggvert_1^infty\=lnfrac{x}{sqrt{x^2+1}}biggvert_1^infty=lnfrac{1}{sqrt{1+1/x^2}}biggvert_1^infty=ln 1-lnfrac{1}{sqrt{2}}=lnsqrt 2$$







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Nov 23 '17 at 7:10









        AndreiAndrei

        11.8k21126




        11.8k21126












        • $begingroup$
          While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
          $endgroup$
          – Dylan
          Nov 23 '17 at 7:33


















        • $begingroup$
          While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
          $endgroup$
          – Dylan
          Nov 23 '17 at 7:33
















        $begingroup$
        While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
        $endgroup$
        – Dylan
        Nov 23 '17 at 7:33




        $begingroup$
        While your answer is correct, I think it'd be more helpful to show OP what was wrong in their method
        $endgroup$
        – Dylan
        Nov 23 '17 at 7:33











        2












        $begingroup$

        First of all:
        $$frac{1}{x(x^2+1)} = frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)$$



        Second of all:



        You have to evaluate the integral $int_1 ^ Rfrac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $infty -infty$ is undetermined.



        $$int_1 ^ Rfrac{1}{x(x^2+1)} = int_1 ^ R left[frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)right]dx$$
        $$=[log |x| - frac{1}2{}log(|x+i|) - frac{1}{2}log(|x-i|)]_1^R$$



        $$=[log |R| - frac{1}2{}log(|R+i|) - frac{1}{2}log(|R-i|)] + frac{1}2{}log(|1+i|) + frac{1}{2}log(|1-i|)]$$



        $$=logleft|frac{R}{(R+i)^{frac{1}{2}}(R-i)^frac{1}{2}}right|+frac{1}{2}log(1+i)(1-i)$$
        $$=logleft|frac{R}{(R^2+1)^frac{1}{2}}right|+frac{1}{2}log(2)$$
        Now taking the limit yields $frac{1}{2}log(2)$.






        share|cite|improve this answer









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          2












          $begingroup$

          First of all:
          $$frac{1}{x(x^2+1)} = frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)$$



          Second of all:



          You have to evaluate the integral $int_1 ^ Rfrac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $infty -infty$ is undetermined.



          $$int_1 ^ Rfrac{1}{x(x^2+1)} = int_1 ^ R left[frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)right]dx$$
          $$=[log |x| - frac{1}2{}log(|x+i|) - frac{1}{2}log(|x-i|)]_1^R$$



          $$=[log |R| - frac{1}2{}log(|R+i|) - frac{1}{2}log(|R-i|)] + frac{1}2{}log(|1+i|) + frac{1}{2}log(|1-i|)]$$



          $$=logleft|frac{R}{(R+i)^{frac{1}{2}}(R-i)^frac{1}{2}}right|+frac{1}{2}log(1+i)(1-i)$$
          $$=logleft|frac{R}{(R^2+1)^frac{1}{2}}right|+frac{1}{2}log(2)$$
          Now taking the limit yields $frac{1}{2}log(2)$.






          share|cite|improve this answer









          $endgroup$
















            2












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            2





            $begingroup$

            First of all:
            $$frac{1}{x(x^2+1)} = frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)$$



            Second of all:



            You have to evaluate the integral $int_1 ^ Rfrac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $infty -infty$ is undetermined.



            $$int_1 ^ Rfrac{1}{x(x^2+1)} = int_1 ^ R left[frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)right]dx$$
            $$=[log |x| - frac{1}2{}log(|x+i|) - frac{1}{2}log(|x-i|)]_1^R$$



            $$=[log |R| - frac{1}2{}log(|R+i|) - frac{1}{2}log(|R-i|)] + frac{1}2{}log(|1+i|) + frac{1}{2}log(|1-i|)]$$



            $$=logleft|frac{R}{(R+i)^{frac{1}{2}}(R-i)^frac{1}{2}}right|+frac{1}{2}log(1+i)(1-i)$$
            $$=logleft|frac{R}{(R^2+1)^frac{1}{2}}right|+frac{1}{2}log(2)$$
            Now taking the limit yields $frac{1}{2}log(2)$.






            share|cite|improve this answer









            $endgroup$



            First of all:
            $$frac{1}{x(x^2+1)} = frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)$$



            Second of all:



            You have to evaluate the integral $int_1 ^ Rfrac{1}{x(x^2+1)}dx$ then take the limit as R goes to infinity and note that $infty -infty$ is undetermined.



            $$int_1 ^ Rfrac{1}{x(x^2+1)} = int_1 ^ R left[frac{1}{x}-frac{1}{2}left(frac{1}{x+i} -frac{1}{x-i}right)right]dx$$
            $$=[log |x| - frac{1}2{}log(|x+i|) - frac{1}{2}log(|x-i|)]_1^R$$



            $$=[log |R| - frac{1}2{}log(|R+i|) - frac{1}{2}log(|R-i|)] + frac{1}2{}log(|1+i|) + frac{1}{2}log(|1-i|)]$$



            $$=logleft|frac{R}{(R+i)^{frac{1}{2}}(R-i)^frac{1}{2}}right|+frac{1}{2}log(1+i)(1-i)$$
            $$=logleft|frac{R}{(R^2+1)^frac{1}{2}}right|+frac{1}{2}log(2)$$
            Now taking the limit yields $frac{1}{2}log(2)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 23 '17 at 7:38









            MereMortal47MereMortal47

            1,274314




            1,274314























                2












                $begingroup$

                put $x = tan theta$ given integral is
                $$int_{frac{pi}{4}}^{frac{pi}{2}} frac{sec^2 theta}{tan theta sec^2 theta} = frac{d(sin theta)}{sin theta} = ln [|sin theta|]^{frac{pi}{2}}_{frac{pi}{4}} = 0-ln frac{1}{sqrt{2}}=ln sqrt{2}$$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  put $x = tan theta$ given integral is
                  $$int_{frac{pi}{4}}^{frac{pi}{2}} frac{sec^2 theta}{tan theta sec^2 theta} = frac{d(sin theta)}{sin theta} = ln [|sin theta|]^{frac{pi}{2}}_{frac{pi}{4}} = 0-ln frac{1}{sqrt{2}}=ln sqrt{2}$$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    put $x = tan theta$ given integral is
                    $$int_{frac{pi}{4}}^{frac{pi}{2}} frac{sec^2 theta}{tan theta sec^2 theta} = frac{d(sin theta)}{sin theta} = ln [|sin theta|]^{frac{pi}{2}}_{frac{pi}{4}} = 0-ln frac{1}{sqrt{2}}=ln sqrt{2}$$






                    share|cite|improve this answer











                    $endgroup$



                    put $x = tan theta$ given integral is
                    $$int_{frac{pi}{4}}^{frac{pi}{2}} frac{sec^2 theta}{tan theta sec^2 theta} = frac{d(sin theta)}{sin theta} = ln [|sin theta|]^{frac{pi}{2}}_{frac{pi}{4}} = 0-ln frac{1}{sqrt{2}}=ln sqrt{2}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 '17 at 8:09









                    Nosrati

                    26.5k62354




                    26.5k62354










                    answered Nov 23 '17 at 6:53









                    MagnetoMagneto

                    886213




                    886213























                        2












                        $begingroup$

                        Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's



                        $$ frac{1}{x(1+x^2)} = frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} $$



                        Then



                        $$ int_1^{infty} f(x) dx = left.left(ln |x| - frac{1}{2}ln |x+i| -
                        frac{1}{2} ln |x-i|right)right|_1^{infty} = left. ln left|frac{x}{sqrt{x^2+1}} right|right|_1^{infty} = frac{1}{2}ln 2 $$



                        Note that
                        $$ lim_{xto infty} frac{x}{sqrt{x^2+1}} = lim_{xtoinfty}frac{1}{sqrt{1+frac{1}{x^2}}} = 1 $$






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 8:09










                        • $begingroup$
                          Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                          $endgroup$
                          – Dylan
                          Nov 23 '17 at 9:08












                        • $begingroup$
                          No problem. Happens to everyone from time to time
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 9:20
















                        2












                        $begingroup$

                        Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's



                        $$ frac{1}{x(1+x^2)} = frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} $$



                        Then



                        $$ int_1^{infty} f(x) dx = left.left(ln |x| - frac{1}{2}ln |x+i| -
                        frac{1}{2} ln |x-i|right)right|_1^{infty} = left. ln left|frac{x}{sqrt{x^2+1}} right|right|_1^{infty} = frac{1}{2}ln 2 $$



                        Note that
                        $$ lim_{xto infty} frac{x}{sqrt{x^2+1}} = lim_{xtoinfty}frac{1}{sqrt{1+frac{1}{x^2}}} = 1 $$






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 8:09










                        • $begingroup$
                          Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                          $endgroup$
                          – Dylan
                          Nov 23 '17 at 9:08












                        • $begingroup$
                          No problem. Happens to everyone from time to time
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 9:20














                        2












                        2








                        2





                        $begingroup$

                        Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's



                        $$ frac{1}{x(1+x^2)} = frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} $$



                        Then



                        $$ int_1^{infty} f(x) dx = left.left(ln |x| - frac{1}{2}ln |x+i| -
                        frac{1}{2} ln |x-i|right)right|_1^{infty} = left. ln left|frac{x}{sqrt{x^2+1}} right|right|_1^{infty} = frac{1}{2}ln 2 $$



                        Note that
                        $$ lim_{xto infty} frac{x}{sqrt{x^2+1}} = lim_{xtoinfty}frac{1}{sqrt{1+frac{1}{x^2}}} = 1 $$






                        share|cite|improve this answer











                        $endgroup$



                        Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's



                        $$ frac{1}{x(1+x^2)} = frac{1}{x} - frac{1}{2(x+i)} - frac{1}{2(x-i)} $$



                        Then



                        $$ int_1^{infty} f(x) dx = left.left(ln |x| - frac{1}{2}ln |x+i| -
                        frac{1}{2} ln |x-i|right)right|_1^{infty} = left. ln left|frac{x}{sqrt{x^2+1}} right|right|_1^{infty} = frac{1}{2}ln 2 $$



                        Note that
                        $$ lim_{xto infty} frac{x}{sqrt{x^2+1}} = lim_{xtoinfty}frac{1}{sqrt{1+frac{1}{x^2}}} = 1 $$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Nov 23 '17 at 9:08

























                        answered Nov 23 '17 at 7:32









                        DylanDylan

                        12.7k31026




                        12.7k31026








                        • 1




                          $begingroup$
                          The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 8:09










                        • $begingroup$
                          Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                          $endgroup$
                          – Dylan
                          Nov 23 '17 at 9:08












                        • $begingroup$
                          No problem. Happens to everyone from time to time
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 9:20














                        • 1




                          $begingroup$
                          The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 8:09










                        • $begingroup$
                          Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                          $endgroup$
                          – Dylan
                          Nov 23 '17 at 9:08












                        • $begingroup$
                          No problem. Happens to everyone from time to time
                          $endgroup$
                          – Andrei
                          Nov 23 '17 at 9:20








                        1




                        1




                        $begingroup$
                        The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                        $endgroup$
                        – Andrei
                        Nov 23 '17 at 8:09




                        $begingroup$
                        The limit at $1$ is not $0$. How can the integral of a positive function be $0$?
                        $endgroup$
                        – Andrei
                        Nov 23 '17 at 8:09












                        $begingroup$
                        Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                        $endgroup$
                        – Dylan
                        Nov 23 '17 at 9:08






                        $begingroup$
                        Thanks. For some reason I thought it evaluates to $ln 1$. I've updated my answer
                        $endgroup$
                        – Dylan
                        Nov 23 '17 at 9:08














                        $begingroup$
                        No problem. Happens to everyone from time to time
                        $endgroup$
                        – Andrei
                        Nov 23 '17 at 9:20




                        $begingroup$
                        No problem. Happens to everyone from time to time
                        $endgroup$
                        – Andrei
                        Nov 23 '17 at 9:20











                        2












                        $begingroup$

                        By enforcing the substitution $x=frac{1}{z}$ the result is straightforward to find:



                        $$ int_{1}^{+infty}frac{dx}{x(x^2+1)}=int_{0}^{1}frac{z,dz}{z^2+1}=left[frac{1}{2}log(1+z^2)right]_{0}^{1}=color{red}{frac{log 2}{2}}.$$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          By enforcing the substitution $x=frac{1}{z}$ the result is straightforward to find:



                          $$ int_{1}^{+infty}frac{dx}{x(x^2+1)}=int_{0}^{1}frac{z,dz}{z^2+1}=left[frac{1}{2}log(1+z^2)right]_{0}^{1}=color{red}{frac{log 2}{2}}.$$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            By enforcing the substitution $x=frac{1}{z}$ the result is straightforward to find:



                            $$ int_{1}^{+infty}frac{dx}{x(x^2+1)}=int_{0}^{1}frac{z,dz}{z^2+1}=left[frac{1}{2}log(1+z^2)right]_{0}^{1}=color{red}{frac{log 2}{2}}.$$






                            share|cite|improve this answer









                            $endgroup$



                            By enforcing the substitution $x=frac{1}{z}$ the result is straightforward to find:



                            $$ int_{1}^{+infty}frac{dx}{x(x^2+1)}=int_{0}^{1}frac{z,dz}{z^2+1}=left[frac{1}{2}log(1+z^2)right]_{0}^{1}=color{red}{frac{log 2}{2}}.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 23 '17 at 19:08









                            Jack D'AurizioJack D'Aurizio

                            289k33281661




                            289k33281661






























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