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I got this question today but I can't see if there is any good way to solve it by hand.

Evaluate the definite integral

$$\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}\,\mathrm{d}x$$

where the series in the numerator and denominator continue infinitely.

If you let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$, solving for $y$ we get $y=\frac{1\pm\sqrt{1+4x}}{2}$. And similarly for the denominator we have $z=\sqrt{xz}$. So $z=x$. So the integral simplifies to

$$\int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

Now my problems are

  1. I don't know what to with the $\pm$.

  2. I tried to solve the integral by separating it as a sum of two fractions. But I can't solve $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

Community
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Soham
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  • See also : https://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation/589293#589293 and https://math.stackexchange.com/questions/322690/convergence-of-nested-radicals – lab bhattacharjee Nov 28 '17 at 10:19
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    "1.I don't know what to with the $\pm$" Use that you know $x\geq 2$ to help you choose one of the solutions. – Arthur Nov 28 '17 at 10:21
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    "without a calculator..." First of all, a calculated value is not a mathematical solution. Second of all, maybe the calculator has an algebraic solver (like Macsyma). Is your real question "does the anti-derivative exist in closed form?" ? – Carl Witthoft Nov 28 '17 at 16:09
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    Note that your technique for simplifying the nested radicals only works if the expressions converge. When the expressions don't converge, it will still give you a value, but that value will be meaningless. Now, in this case, you can show that both sequences are increasing and bounded above, and so must converge. But in general, you should be very cautious with the "$y = f(x, y)$, so $y = f(x, f(x, f(x, ...)))$" concept. – Paul Sinclair Nov 28 '17 at 17:32
  • @CarlWitthoft I agree with you but I would have liked that better if it was an indefinite integral.;-) – Soham Nov 29 '17 at 03:25

4 Answers4

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As you did, let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$. Clearly $y\ge0$ and $y$ satisfies $$ y^2-y-x=0 $$ from which you have $$ y=\frac{1\pm\sqrt{4x+1}}{2}. $$ Since $x\in[2,12]$ and $y\ge0$, you must choose "$+$". Since if you choose "$-$", then $$ y=\frac{1-\sqrt{4x+1}}{2}<0. $$ Now, under $t=\sqrt{1+4x}$ \begin{eqnarray} &&\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}dx\\ &=&\int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx\\ &=&\int_2^{12}\frac{1}{2x}dx+\int_2^{12}\frac{\sqrt{1+4x}}{2x}dx\\ &=&\frac12\ln x\bigg|_2^{12}+\int_3^7\frac{t^2}{t^2-1}dt\\ &=&\frac12\ln6+\bigg(t+\frac12\ln\frac{t-1}{t+1}\bigg)\bigg|_3^7\\ &=&\frac12\ln6+4+\frac12\ln\frac32\\ &=&\ln3+4. \end{eqnarray}

xpaul
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7

Hint:

 1.

For real $y,$ $\sqrt{y^2}=|y|\ge0$

and $\sqrt{1+4x}\ge3$ for $2\le x\le12\implies1-\sqrt{1+4x}<0$

2.

Set $\sqrt{1+4x}=u\implies4x=u^2-1$

lab bhattacharjee
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$ \int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}dx$

Now, notice that the limits are defined for positive values of x. Also,

$y=\frac{1\pm\sqrt{1+4x}}{2}<0$ if you consider the negative sign for $x>0$.

But, $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}>0$ since the square root of a number is always positive for it to be a function (otherwise, for only one value of $x$, there'll be two values of y i.e., multiple mapping into range for only one value in the domain).

So, $y>0$.

So, $I= \int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx$

Now, just substitute $4x+1=u^2$ (as mentioned in one of the above answers).

$\implies I=\int \frac{u}{u-1}du$ with appropriate limits

Your IDE
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1

For 2. notice that

$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx=\int_{2}^{12}\frac{1+4x}{2x\sqrt{1+4x}}=\frac{1}{2}\left(\int_{2}^{12}\frac{1}{x\sqrt{1+4x}}dx+4\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}\right).$$

Now the second integral is pretty easy

$$2\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}=\left[\sqrt{1+4x}\right]_{2}^{12}=4,$$

for the first one instead

$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{2}^{12}\frac{dx}{x\sqrt{1+\left(2\sqrt{x}\right)^2}},$$

let $2\sqrt{x}=\sinh y$, so $x=\frac{\sinh^2 y}{4}$ and $dx=\frac{\cosh y \sinh y}{2}\ dy$. Hence

$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{\frac{\cosh y \sinh y}{2}}{\frac{\sinh^2}{4} y\sqrt{1+\sinh^2 y}}dy=\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{dy}{\sinh y}=$$

$$=2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy.$$

Now let $e^y=z$, so $e^y\ dy=dz$

$$2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy=2\int_{\eta}^{\psi}\frac{dz}{z^2-1},$$

where $\eta=e^{\sinh^{-1}{2\sqrt2}}$ and $\psi={e^{\sinh^{-1}{4\sqrt3}}}.$

$$2\int_{\eta}^{\psi}\frac{dz}{z^2-1}=\ln\left|\frac{z-1}{z+1}\right|_{\eta}^{\psi}=\ln\left|\frac{e^y-1}{e^y+1}\right|_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}=$$

$$=\ln\left|\frac{e^{\sinh^{-1}{2\sqrt{x}}}-1}{e^{\sinh^{-1}{2\sqrt{x}}}+1}\right|_{2}^{12},$$

so

$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$

Frieren
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