# Thread: an integral from the 2010 MIT Integration Bee

1. ## an integral from the 2010 MIT Integration Bee

The following integral was on the qualifying test:

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx$

Any ideas? I'm stumped.

2. Originally Posted by Random Variable
The following integral was on the qualifying test:

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx$

Any ideas? I'm stumped.
Let $x=\sec^2(z)$ or $x=\sinh^2(z)$?

let $y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}$

then $y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy$

solving for y

$y = \frac{x}{2} \pm \frac{\sqrt{x^{2}+4}}{2}$

then using the positive root since we're integrating over positive values of x

$\int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx \ = \int^{1}_{0} \Big(\frac{x}{2} + \frac{\sqrt{x^{2}+4}}{2} \Big) \ dx$ ?

4. then you solve for y using quadratic formula ?

y= x/2 + sqrt(x^2 +4)/2 ?

or am I way off base?

5. Originally Posted by Random Variable

let $y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}}$

then $y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy$ ?
Originally Posted by Calculus26
then you solve for y using quadratic formula ?

y= x/2 + sqrt(x^2 +4)/2 ?

or am I way off base?
No no! That is correct. I didn't realize it was supposed to be an infinite surd . Yes, that method works...if you can prove that the surd converges .

P.S. You can choose the $\pm$ according to the fact that $y>0$ for $x\in[0,1]$

6. Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?

7. Originally Posted by Calculus26
Then since y > 0 we take y= x/2 + sqrt(x^2 +4)/2 ?
I would agree with that. the answer is $\frac{1+\sqrt{5}}{4}+\text{arccsch}\left(2\right)$