I hope there will be no convergence issues (eighteenth integral)

Calculus Level 5

∫ 0 1 ( x + x x + x x + x x + . . . ) d x \displaystyle {\huge \int_{0}^{1} } \left( x + \dfrac{x}{x+ \dfrac{x}{x+\dfrac{x}{x+...}}} \right) \, dx

If the above integral can be represented in the form

a b + c d f − g ln ⁡ ( h + j k ) \dfrac{a}{b} + \dfrac{c \sqrt{d}}{f} - g \ln \left( \dfrac{h + \sqrt{j}}{k}\right)

where

  • gcd ⁡ ( a , b ) = gcd ⁡ ( c , f ) = gcd ⁡ ( h , k ) = 1 \gcd(a, b) = \gcd(c, f) = \gcd(h, k) = 1
  • a , b , c , d , f , g , h , j , k a, b, c, d, f, g, h, j, k are all positive integers
  • d , j d, j are square-free
  • g = 2 g = 2

then find a + b + c + d + f + g + h + j + k a+b+c+d+f+g+h+j+k .

Bonus: What special number is contained somewhere in the answer to this problem?


The answer is 27.

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Hobart Pao by Hobart Pao , 23, USA

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2 solutions

Let I = ∫ 0 1 ( x + x x + x x + x x + . . . ) d x \displaystyle {\huge \int_{0}^{1} } \left( x + \dfrac{x}{x+ \dfrac{x}{x+\dfrac{x}{x+...}}} \right) \, dx

Let the expression given under integral be A. So,

x + x A = A \large x + \frac{x}{A} = A

Solving for A we get ,

A = x + ( x + 2 ) 2 − 2 2 2 \large \displaystyle A = \frac{x + \sqrt{(x+2)^{2} - 2^{2}}}{2}

I = 1 2 ∫ 0 1 x + ( x + 2 ) 2 − 2 2 d x \large I = \displaystyle \frac{1}{2} \int_{0}^{1} x + \sqrt{(x+2)^{2} - 2^{2}} dx

For the part ( x + 2 ) 2 − 2 2 \sqrt{(x+2)^{2} - 2^{2}} we get the following after integration ,

I = 1 2 [ x 2 2 ] 0 1 + [ 1 2 ( x + 2 ) ( x + 2 ) 2 − 2 2 − 1 2 2 2 l o g ∣ ( x + 2 ) + ( x + 2 ) 2 − 2 2 ∣ ] 0 1 \large \displaystyle I = \frac{1}{2} [\frac{x^{2}}{2}]_{0}^{1} + [\frac{1}{2}(x+2)\sqrt{(x+2)^{2} - 2^{2}} - \frac{1}{2}2^{2}log| (x + 2) + \sqrt{(x+2)^{2} - 2^{2}}|]_{0}^{1} [Proof for this formulae is skipped as it is a general one]

I = 1 2 [ 1 2 + 3 ( 5 ) 2 − 2 [ l o g ∣ 3 + ( 5 ) ∣ − l o g 2 ] \large \displaystyle I = \frac{1}{2} [\frac{1}{2} + \frac{3\sqrt(5)}{2} - 2[log|3 + \sqrt(5)| - log2]

I = 1 2 [ 1 2 + 3 ( 5 ) 2 − 2 l o g ∣ 3 + ( 5 ) 2 ∣ ] \large \displaystyle I = \frac{1}{2} [\frac{1}{2} + \frac{3\sqrt(5)}{2} - 2log|\frac{3+\sqrt(5)}{2}|]

I = 1 4 + 3 ( 5 ) 4 − 2 l o g ∣ 3 + ( 5 ) 2 ∣ \large \displaystyle I = \frac{1}{4} + \frac{3\sqrt(5)}{4} - 2log|\sqrt{\frac{3+\sqrt(5)}{2}}|

I = 1 4 + 3 ( 5 ) 4 − 2 l o g ∣ 5 + 1 + 2 ( 5 ) 4 ∣ \large \displaystyle I = \frac{1}{4} + \frac{3\sqrt(5)}{4} - 2log|\sqrt{\frac{5+1+2\sqrt(5)}{4}}|

I = 1 4 + 3 ( 5 ) 4 − 2 l o g ∣ ( ( 5 ) + 1 2 ) 2 ∣ \large \displaystyle I = \frac{1}{4} + \frac{3\sqrt(5)}{4} - 2log|\sqrt{(\frac{\sqrt(5) + 1}{2})^{2}}|

I = 1 4 + 3 ( 5 ) 4 − 2 l o g ∣ 1 + ( 5 ) 2 ∣ \large \displaystyle I = \frac{1}{4} + \frac{3\sqrt(5)}{4} - 2log|\frac{1+\sqrt(5)}{2}|

Therefore , a + b + c + d + f + g + h + j + k = 27

Bonus : It is the Golden ration inside the logarithm.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

There are indeed convergence issues. You have to explain why

  1. The expression converges to a finite value.
  2. The expression evaluates to the positive root.

For the bonus question, the golden ratio!

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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