I want to calculate the antiderivative of this function .

$$ y(x) = \sqrt{x+\left(\sqrt{{x}+\sqrt{x+\cdots}}\right)}$$


let $y = \sqrt{x + y}$ then: $\displaystyle\int {ydy=\displaystyle\frac{1}{2\sqrt{x+y}}}+c =\frac{1}{2\sqrt{x+\sqrt{x+\left(\sqrt{{x}+\sqrt{x+\cdots}}\right)}}}+c$ or it must to integrate with respect to $x$ ?

  • $\begingroup$ Hint: Try to find a simpler expression for $y$ before you integrate. You need to be a bit careful with the convergence. $\endgroup$ – mickep Feb 7 '18 at 13:27
  • $\begingroup$ related math.stackexchange.com/questions/2480022/… $\endgroup$ – Guy Fsone Feb 7 '18 at 13:28
  • $\begingroup$ @Jack Wonder What do you want to integrate it with ? y or x? $\endgroup$ – Deepak M S Feb 7 '18 at 13:30
  • $\begingroup$ I want to integrate for x as shown in the title $\endgroup$ – user517526 Feb 7 '18 at 13:31
  • 1
    $\begingroup$ If the differential element in the integral is $dx$ then integrate with respect to $x$, by converting all $y$ into $x$. Look, whatever is the variable of integration, in this case which is $x$, there shouldn't be another variable present which depends on previous, which is $y$. If you want to integrate w.r.t $y$, surely you can but you need to change the varibles accordingly. You will have to find relation between $dx$ and $dy$ and the properly replace $dx$ with $dy$, remove $x$ completely from the expression, and then you are all set to integrate w.r.t. $y$. $\endgroup$ – Jaideep Khare Feb 7 '18 at 13:31

Thanks to some fruitful discussion with @MPW4 your initial expression makes any sense before hand in $\Bbb R$ only if $x>0$

From now we assume $x>0.$ otherwise $y(x)$ might clearly be undefined as well. we have $$y = \sqrt{x+\left(\sqrt{{x}+\sqrt{x+\cdots}}\right)}\implies y(x) = \sqrt{x+y(x)}~~~~y(x)\ge 0\\ \implies y^2(x)-y(x)-x=0, ~~~~y(x)\ge 0 \\\implies y(x)\equiv y_\pm(x) = \frac{1\pm\sqrt{1+4x}}{2}~~~~y(x)\ge 0$$

Hence you have to choose one branch, either $y_-$ or $y_+$ but using the constraints, $x\ge 0~~$ and $~~y(x)\ge 0$ show that the only possible value is given by, $$ y(x)= y_+ = \frac{1+\sqrt{1+4x}}{2}$$

since $$y_-(x)\ge 0\implies x\le 0$$

Conclusion we have $ y(x)= y_+ = \frac{1+\sqrt{1+4x}}{2}$ and $$\int y(x)dx = \frac{x+\frac{1}{6}(1+4x)^{3/2}}{2}$$


But rather you could consider $x\equiv x(y) $ as function of $y$ variable then, you obviously get $$ \int x(y)dy = \int y^2-ydy=\frac{1}{3}y^3-\frac{1}{2}y^2$$

  • $\begingroup$ Your conclusion is incorrect. There may be two values of $y$ which satisfy your derived quadratic, but that doesn't mean they are both the limiting value (for a given $x$, $y$ is defined as a limit, after all). You have simply introduced an extraneous root by squaring both sides. The only reasons $y(x)$ could be undefined are (1) some term in the sequence is undefined (e.g., square root of a negative number) and (2) the sequence diverges. Your reasoning applied to the function $y=x$ would produce $y^2(x) = x^2$ and then $y(x)=\pm\sqrt{x^2} = \pm|x|$. $\endgroup$ – MPW Feb 7 '18 at 13:48
  • 1
    $\begingroup$ @zeraouliarafik the OP you should decide to choose one branch functions I proposed otherwise it is not a function. I think in such cases it is baby analysis $\endgroup$ – Guy Fsone Feb 7 '18 at 14:13
  • 1
    $\begingroup$ You can't just "pick a branch". For positive $x$, only the "$+$" branch could possibly be the actual value of $y$. If you choose the "$-$" branch, then the resulting $y$ is negative (the radicand is greater than $1$, so the difference is negative). This is clearly impossible since you are just taking a sequence of square roots of positive numbers. As I mentioned, YOU have introduced the ambiguous sign by squaring. There AREN'T two branches. $\endgroup$ – MPW Feb 7 '18 at 14:19
  • 1
    $\begingroup$ When $x<0$, the function is undefined, assuming we're working in the real domain. For example, the value of $y$ when $x=-\tfrac19$ would have to be the limit of the sequence $$\sqrt{-\tfrac19},\sqrt{-\tfrac19+\sqrt{-\tfrac19}},\sqrt{-\tfrac19+\sqrt{-\tfrac19+\sqrt{-\tfrac19}}},\ldots$$ but non of these terms are defined. $\endgroup$ – MPW Feb 7 '18 at 14:32
  • 1
    $\begingroup$ @MPW yeaaar that could be a good an actually it is a surprising result. you start with complex and end with real number $\endgroup$ – Guy Fsone Feb 7 '18 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy