# Can this definite integral involving series be solved without a calculator?

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

• Nov 28, 2017 at 10:19
• "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. Nov 28, 2017 at 10:21
• "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?" ? Nov 28, 2017 at 16:09
• 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. Nov 28, 2017 at 17:32
• @CarlWitthoft I agree with you but I would have liked that better if it was an indefinite integral.;-) Nov 29, 2017 at 3:25

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}

• Can you explain your substitution of t in the second integral? Nov 29, 2017 at 3:18
• @tatan - if $t = \sqrt{1+4x}$, then $x = \frac{t^2 + 1}4$, so $dx = \frac t2 dt$. Nov 29, 2017 at 3:37
• @PaulSinclair Thanks Nov 29, 2017 at 3:43
• correction: $x = \frac{t^2 -1}4$. Nov 29, 2017 at 3:44

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

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

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

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

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

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

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

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