an integral from the 2010 MIT Integration Bee

**Random Variable** · Jan 28th 2010, 08:48 PM

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.

**Drexel28** · Jan 28th 2010, 08:51 PM

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 $x=\sinh^2(z)$ ?

**Random Variable** · Jan 28th 2010, 10:01 PM

How about the following:

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

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

**Calculus26** · Jan 28th 2010, 10:07 PM

then you solve for y using quadratic formula ?

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

or am I way off base?

**Drexel28** · Jan 28th 2010, 10:09 PM

Originally Posted by Random Variable

What about the following:

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 $x\in[0,1]$

**Calculus26** · Jan 28th 2010, 10:11 PM

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

**Drexel28** · Jan 28th 2010, 10:15 PM

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

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