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












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











$endgroup$








  • 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







share|cite|improve this question











$endgroup$








  • 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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
5






active

oldest

votes


















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












$begingroup$

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












    $begingroup$

    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












      $begingroup$

      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












      $begingroup$

      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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2533478%2fevaluating-int-1-infty-frac1xx21-dx%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        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












        share|cite|improve this answer



        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









        $endgroup$


















          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












            2








            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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2533478%2fevaluating-int-1-infty-frac1xx21-dx%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Ellipse (mathématiques)

                                Image
                                Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                                Read more

                                Quarter-circle Tiles

                                Image
                                up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                                Read more

                                Mont Emei

                                Image
                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                                Read more