Thread: an integral from the 2010 MIT Integration Bee

The following integral was on the qualifying test:

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

Any ideas? I'm stumped.

Originally Posted by Random Variable

The following integral was on the qualifying test:

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

Any ideas? I'm stumped.

Let $\displaystyle x=\sec^2(z)$ or $\displaystyle x=\sinh^2(z)$?

How about the following:

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

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

solving for y

$\displaystyle 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

$\displaystyle \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 $ ?

then you solve for y using quadratic formula ?

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

or am I way off base?

Originally Posted by Random Variable

What about the following:

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

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

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 $\displaystyle \pm$ according to the fact that $\displaystyle y>0$ for $\displaystyle x\in[0,1]$

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

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 $\displaystyle \frac{1+\sqrt{5}}{4}+\text{arccsch}\left(2\right)$