A calculus problem by Jose Sacramento

Calculus Level 3

Let f : [ 0 , 1 ] → R f: [0,1] \to \mathbb R be a continuous and positive function satisfying f ( x ) f ( 1 − x ) = 1 f(x) f(1-x) = 1 for 0 ≤ x ≤ 1 0 \leq x \leq 1 .

Compute the closed form of ∫ 0 1 1 1 + f ( x ) d x \displaystyle \int_0^1 \dfrac1{1+f(x)} \, dx .


The answer is 0.5.

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Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

Chew-Seong Cheong
Oct 18, 2016

I = ∫ 0 1 1 1 + f ( x ) d x Using identity ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 1 2 ∫ 0 1 1 1 + f ( x ) + 1 1 + f ( 1 − x ) d x = 1 2 ∫ 0 1 1 + f ( 1 − x ) + 1 + f ( x ) ( 1 + f ( x ) ( 1 + f ( 1 − x ) ) d x = 1 2 ∫ 0 1 2 + f ( x ) + f ( 1 − x ) 1 + f ( x ) + f ( 1 − x ) + f ( x ) f ( 1 − x ) d x Given that f ( x ) f ( 1 − x ) = 1 = 1 2 ∫ 0 1 2 + f ( x ) + f ( 1 − x ) 2 + f ( x ) + f ( 1 − x ) d x = 1 2 ∫ 0 1 d x = 1 2 = 0.5 \begin{aligned} I & = \int_0^1 \frac 1{1+f(x)} dx \quad \quad \small \color{#3D99F6}{\text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^1 \frac 1{1+f(x)} + \frac 1{1+f(1-x)} dx \\ & = \frac 12 \int_0^1 \frac {1+f(1-x) + 1+f(x)}{(1+f(x)(1+f(1-x))} dx \\ & = \frac 12 \int_0^1 \frac {2+f(x)+f(1-x)}{1+f(x)+f(1-x)+\color{#3D99F6}{f(x)f(1-x)}} dx \quad \quad \small \color{#3D99F6}{\text{Given that }f(x)f(1-x) = 1} \\ & = \frac 12 \int_0^1 \frac {2+f(x)+f(1-x)}{{\color{#3D99F6}{2}}+f(x)+f(1-x)} dx \\ & = \frac 12 \int_0^1 \ dx = \frac 12 = \boxed{0.5} \end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

As an aside, your solution starts off by making the assumption that I I exists and is finite. How can we justify that?

Hint: You have not used all the conditions of the question as yet.
E.g. the function f ( x ) = − 1 f(x) = - 1 satisfies the functional equation, but doesn't have a finite I I .

Calvin Lin Staff - 4 years, 8 months ago

Wonderfull solution... Thank you. Cheers

Jose Sacramento - 4 years, 8 months ago

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