Square root of fractional part integral












2












$begingroup$


Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$



Where ${x}$ denotes the fractional part of $x$.






integration zeta-functions fractional-part







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Have you tried any methods of evaluating it? Perhaps in series form?
    $endgroup$
    – B. Mehta
    Jul 19 '18 at 16:31
















2












$begingroup$


Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$



Where ${x}$ denotes the fractional part of $x$.






integration zeta-functions fractional-part







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Have you tried any methods of evaluating it? Perhaps in series form?
    $endgroup$
    – B. Mehta
    Jul 19 '18 at 16:31














2












2








2





$begingroup$


Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$



Where ${x}$ denotes the fractional part of $x$.






integration zeta-functions fractional-part







share|cite|improve this question











$endgroup$




Does the following integral have a closed form ?
$$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx$$



Where ${x}$ denotes the fractional part of $x$.






integration zeta-functions fractional-part




integration zeta-functions fractional-part






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 17:38









YuiTo Cheng

2,1863937




2,1863937










asked Jul 19 '18 at 16:26









Kays Tomy Kays Tomy

19717




19717








  • 6




    $begingroup$
    Have you tried any methods of evaluating it? Perhaps in series form?
    $endgroup$
    – B. Mehta
    Jul 19 '18 at 16:31














  • 6




    $begingroup$
    Have you tried any methods of evaluating it? Perhaps in series form?
    $endgroup$
    – B. Mehta
    Jul 19 '18 at 16:31








6




6




$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31




$begingroup$
Have you tried any methods of evaluating it? Perhaps in series form?
$endgroup$
– B. Mehta
Jul 19 '18 at 16:31










3 Answers
3






active

oldest

votes


















4












$begingroup$

Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.



The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using



$$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$



From this, the area under a tooth is given by



$$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$



The total area is the sum of the $I_n$, which doesn't seem to have a closed form.



enter image description here



If we use the Taylor development of the arc tangent,



$$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$



Now if we sum on $n$, noticing that the first term will telescope with the last one, we get



$$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the same solution as that of Marty Cohen, but explained my way.
    $endgroup$
    – Yves Daoust
    Jul 19 '18 at 21:00





















2












$begingroup$

You want
$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
$.



Consider



$I
=int_{0}^{1}f({1/x})dx
$
where
$f(0) = 0.
f(1) = 1,
f'(x) ge 0
$.



Since
$n le 1/x
le n+1
$
for
$1/(n+1)
le x
le 1/n
$,



$begin{array}\
I
&=int_{0}^{1}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
&=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
&=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
qquad y=1/x, x=1/y, dx=-dy/y^2\
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
end{array}
$



so your result is



$begin{array}
I
&=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
&=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
&=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
qquadtext{(according to Wolfy)}\
end{array}
$



Note:
You can do this
for any function
(instead of $1/x$)
such that
$g(1) = 1,
g'(x) > 0,
g(x) to infty$
as $x to 0$.
You have to look
at $g^{(-1)}(x)$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
    , dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
    ^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$



    With the aid of computer algebra, we obtain integral and approximate series with value.



    Probably closed form of the sum does not exist.




    Supplement:



    Borrowing sum from user: Yves Daoust we can write:



    $$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$



    where: $psi left(1+xright)$ is PolyGamma function






    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%2f2856812%2fsquare-root-of-fractional-part-integral%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.



      The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using



      $$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$



      From this, the area under a tooth is given by



      $$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$



      The total area is the sum of the $I_n$, which doesn't seem to have a closed form.



      enter image description here



      If we use the Taylor development of the arc tangent,



      $$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$



      Now if we sum on $n$, noticing that the first term will telescope with the last one, we get



      $$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is the same solution as that of Marty Cohen, but explained my way.
        $endgroup$
        – Yves Daoust
        Jul 19 '18 at 21:00


















      4












      $begingroup$

      Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.



      The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using



      $$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$



      From this, the area under a tooth is given by



      $$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$



      The total area is the sum of the $I_n$, which doesn't seem to have a closed form.



      enter image description here



      If we use the Taylor development of the arc tangent,



      $$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$



      Now if we sum on $n$, noticing that the first term will telescope with the last one, we get



      $$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        This is the same solution as that of Marty Cohen, but explained my way.
        $endgroup$
        – Yves Daoust
        Jul 19 '18 at 21:00
















      4












      4








      4





      $begingroup$

      Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.



      The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using



      $$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$



      From this, the area under a tooth is given by



      $$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$



      The total area is the sum of the $I_n$, which doesn't seem to have a closed form.



      enter image description here



      If we use the Taylor development of the arc tangent,



      $$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$



      Now if we sum on $n$, noticing that the first term will telescope with the last one, we get



      $$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$






      share|cite|improve this answer











      $endgroup$



      Due to the complexity of the computation (see other answers), we consider integrating $x,dy$ instead of $y,dx$.



      The function has discontinuities at every $x=dfrac1n$, which are jumps from $y=0$ to $y=1$. We can invert the function between $dfrac1{n+1}$ and $dfrac1n$ using



      $$y=sqrt{frac1x-n}iff x=frac1{y^2+n}.$$



      From this, the area under a tooth is given by



      $$I_n=int_0^1frac{dy}{y^2+n}-frac1{n+1}=frac1{sqrt n}arctanfrac1{sqrt n}-frac1{n+1}.$$



      The total area is the sum of the $I_n$, which doesn't seem to have a closed form.



      enter image description here



      If we use the Taylor development of the arc tangent,



      $$frac1{sqrt n}left(frac1{sqrt n}-frac1{3nsqrt n}+frac1{5n^2sqrt n}-cdotsright)-frac1{n+1}.$$



      Now if we sum on $n$, noticing that the first term will telescope with the last one, we get



      $$1-frac13zeta(2)+frac15zeta(3)-frac17zeta(4)+cdots$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jul 19 '18 at 21:09

























      answered Jul 19 '18 at 20:59









      Yves DaoustYves Daoust

      131k676229




      131k676229












      • $begingroup$
        This is the same solution as that of Marty Cohen, but explained my way.
        $endgroup$
        – Yves Daoust
        Jul 19 '18 at 21:00




















      • $begingroup$
        This is the same solution as that of Marty Cohen, but explained my way.
        $endgroup$
        – Yves Daoust
        Jul 19 '18 at 21:00


















      $begingroup$
      This is the same solution as that of Marty Cohen, but explained my way.
      $endgroup$
      – Yves Daoust
      Jul 19 '18 at 21:00






      $begingroup$
      This is the same solution as that of Marty Cohen, but explained my way.
      $endgroup$
      – Yves Daoust
      Jul 19 '18 at 21:00













      2












      $begingroup$

      You want
      $int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
      $.



      Consider



      $I
      =int_{0}^{1}f({1/x})dx
      $
      where
      $f(0) = 0.
      f(1) = 1,
      f'(x) ge 0
      $.



      Since
      $n le 1/x
      le n+1
      $
      for
      $1/(n+1)
      le x
      le 1/n
      $,



      $begin{array}\
      I
      &=int_{0}^{1}f({1/x})dx\
      &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
      &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
      &=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
      qquad y=1/x, x=1/y, dx=-dy/y^2\
      &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
      &=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
      end{array}
      $



      so your result is



      $begin{array}
      I
      &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
      &=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
      &=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
      qquadtext{(according to Wolfy)}\
      end{array}
      $



      Note:
      You can do this
      for any function
      (instead of $1/x$)
      such that
      $g(1) = 1,
      g'(x) > 0,
      g(x) to infty$
      as $x to 0$.
      You have to look
      at $g^{(-1)}(x)$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You want
        $int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
        $.



        Consider



        $I
        =int_{0}^{1}f({1/x})dx
        $
        where
        $f(0) = 0.
        f(1) = 1,
        f'(x) ge 0
        $.



        Since
        $n le 1/x
        le n+1
        $
        for
        $1/(n+1)
        le x
        le 1/n
        $,



        $begin{array}\
        I
        &=int_{0}^{1}f({1/x})dx\
        &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
        &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
        &=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
        qquad y=1/x, x=1/y, dx=-dy/y^2\
        &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
        &=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
        end{array}
        $



        so your result is



        $begin{array}
        I
        &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
        &=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
        &=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
        qquadtext{(according to Wolfy)}\
        end{array}
        $



        Note:
        You can do this
        for any function
        (instead of $1/x$)
        such that
        $g(1) = 1,
        g'(x) > 0,
        g(x) to infty$
        as $x to 0$.
        You have to look
        at $g^{(-1)}(x)$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You want
          $int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
          $.



          Consider



          $I
          =int_{0}^{1}f({1/x})dx
          $
          where
          $f(0) = 0.
          f(1) = 1,
          f'(x) ge 0
          $.



          Since
          $n le 1/x
          le n+1
          $
          for
          $1/(n+1)
          le x
          le 1/n
          $,



          $begin{array}\
          I
          &=int_{0}^{1}f({1/x})dx\
          &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
          &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
          &=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
          qquad y=1/x, x=1/y, dx=-dy/y^2\
          &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
          &=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
          end{array}
          $



          so your result is



          $begin{array}
          I
          &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
          &=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
          &=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
          qquadtext{(according to Wolfy)}\
          end{array}
          $



          Note:
          You can do this
          for any function
          (instead of $1/x$)
          such that
          $g(1) = 1,
          g'(x) > 0,
          g(x) to infty$
          as $x to 0$.
          You have to look
          at $g^{(-1)}(x)$.






          share|cite|improve this answer









          $endgroup$



          You want
          $int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx
          $.



          Consider



          $I
          =int_{0}^{1}f({1/x})dx
          $
          where
          $f(0) = 0.
          f(1) = 1,
          f'(x) ge 0
          $.



          Since
          $n le 1/x
          le n+1
          $
          for
          $1/(n+1)
          le x
          le 1/n
          $,



          $begin{array}\
          I
          &=int_{0}^{1}f({1/x})dx\
          &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f({1/x})dx\
          &=sum_{n=1}^{infty}int_{1/(n+1)}^{1/n}f(1/x-n)dx\
          &=sum_{n=1}^{infty}int_{n}^{n+1}f(y-n)dy/y^2
          qquad y=1/x, x=1/y, dx=-dy/y^2\
          &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
          &=int_{0}^{1}f(y)dysum_{n=1}^{infty}dfrac1{(y+n)^2}\
          end{array}
          $



          so your result is



          $begin{array}
          I
          &=sum_{n=1}^{infty}int_{0}^{1}f(y)dy/(y+n)^2\
          &=sum_{n=1}^{infty}int_{0}^{1}sqrt{y}dy/(y+n)^2\
          &=sum_{n=1}^{infty}left(dfrac{cot(sqrt{n})}{sqrt{n}} - dfrac1{n + 1}right)
          qquadtext{(according to Wolfy)}\
          end{array}
          $



          Note:
          You can do this
          for any function
          (instead of $1/x$)
          such that
          $g(1) = 1,
          g'(x) > 0,
          g(x) to infty$
          as $x to 0$.
          You have to look
          at $g^{(-1)}(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 19 '18 at 18:56









          marty cohenmarty cohen

          74.9k549130




          74.9k549130























              2












              $begingroup$

              $$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
              , dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
              ^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$



              With the aid of computer algebra, we obtain integral and approximate series with value.



              Probably closed form of the sum does not exist.




              Supplement:



              Borrowing sum from user: Yves Daoust we can write:



              $$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$



              where: $psi left(1+xright)$ is PolyGamma function






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                $$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
                , dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
                ^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$



                With the aid of computer algebra, we obtain integral and approximate series with value.



                Probably closed form of the sum does not exist.




                Supplement:



                Borrowing sum from user: Yves Daoust we can write:



                $$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$



                where: $psi left(1+xright)$ is PolyGamma function






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
                  , dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
                  ^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$



                  With the aid of computer algebra, we obtain integral and approximate series with value.



                  Probably closed form of the sum does not exist.




                  Supplement:



                  Borrowing sum from user: Yves Daoust we can write:



                  $$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$



                  where: $psi left(1+xright)$ is PolyGamma function






                  share|cite|improve this answer











                  $endgroup$



                  $$int_{0}^{1}sqrt{bigg{frac{1}{x}bigg}}dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}}sqrt{bigg{frac{1}{x}bigg}}
                  , dx=\sum _{n=1}^{infty } int_{frac{1}{n+1}}^{frac{1}{n}} sqrt{frac{1}{x}-n} , dx=\sum _{n=1}^{infty } frac{-4 sqrt{n}+pi +n pi -2 (1+n) sin
                  ^{-1}left(frac{-1+n}{1+n}right)}{4 sqrt{n} (1+n)}$$



                  With the aid of computer algebra, we obtain integral and approximate series with value.



                  Probably closed form of the sum does not exist.




                  Supplement:



                  Borrowing sum from user: Yves Daoust we can write:



                  $$sum _{n=1}^{infty } left(frac{cot ^{-1}left(sqrt{n}right)}{sqrt{n}}-frac{1}{n+1}right)=1-gamma -int_0^1 frac{psi (1+x)}{2 sqrt{x}} , dxapprox 0.60204991$$



                  where: $psi left(1+xright)$ is PolyGamma function







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 20 '18 at 7:29

























                  answered Jul 19 '18 at 20:23









                  Mariusz IwaniukMariusz Iwaniuk

                  1,9771718




                  1,9771718






























                      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%2f2856812%2fsquare-root-of-fractional-part-integral%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

                      'app-layout' is not a known element: how to share Component with different Modules

                      Image
                      0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                      Read more

                      android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                      Read more

                      WPF add header to Image with URL pettitions [duplicate]

                      Image
                      up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                      Read more