Simplify and Substitute

Calculus Level 5

∫ 0 ∞ d x x [ x 2 + ( 1 + 2 2 ) x + 1 ] [ 1 − x + x 2 − x 3 + ⋯ + x 50 ] \large \int_0^{\infty} \frac{dx}{\sqrt{x}\left[x^2 + \left(1 + 2\sqrt{2}\right)x + 1\right]\left[1 - x + x^2 - x^3 + \cdots + x^{50}\right]}

The value of the above integral can be expressed in the form π ( a b − c ) \pi\left(a\sqrt{b} - c\right) , where a a , b b , and c c are coprime positive integers and b b is square-free. Find ( a + 1 ) ( b + 3 ) ( c + 5 ) (a + 1)(b + 3)(c + 5) .

Bonus: Generalize for integrals of the form

∫ 0 ∞ d x x [ x 2 + a x + 1 ] [ 1 − x + x 2 − x 3 + ⋯ + ( − x ) n ] \int_0^{\infty} \frac{dx}{\sqrt{x}\left[x^2 + ax + 1\right]\left[1 - x + x^2 - x^3 + \cdots + (- x)^n\right]}


The answer is 60.

Zach Abueg by Zach Abueg , 22, USA

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1 solution

Chew-Seong Cheong
Aug 24, 2017

I ( a ) = ∫ 0 ∞ d x x ( x 2 + a x + 1 ) ( 1 − x + x 2 + ⋯ + ( − x ) n ) = ∫ 0 ∞ 1 + x x ( x 2 + a x + 1 ) ( 1 + ( − 1 ) n x n + 1 ) d x . . . ( 1 ) \begin{aligned} I(a) & = \int_0^\infty \frac {dx}{\sqrt x \left(x^2+ax+1\right)\color{#3D99F6}\left(1-x+x^2+\cdots+(-x)^n\right)} \\ & = \int_0^\infty \frac {\color{#3D99F6}1+x}{\sqrt x \left(x^2+ax+1\right)\color{#3D99F6}\left(1+(-1)^nx^{n+1}\right)} dx & ...(1) \end{aligned}

Using the identity ∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ f ( 1 x ) x 2 d x \displaystyle \int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} \ dx ,

I ( a ) = ∫ 0 ∞ 1 + 1 x x 2 x ( 1 x 2 + a x + 1 ) ( 1 + 1 ( − 1 ) n x n + 1 ) d x = ∫ 0 ∞ ( 1 + x ) ( − 1 ) n x n + 1 x ( x 2 + a x + 1 ) ( 1 + ( − 1 ) n x n + 1 ) d x . . . ( 2 ) \begin{aligned} I(a) & = \int_0^\infty \frac {1+\frac 1x}{\frac {x^2}{\sqrt x} \left(\frac 1{x^2}+ \frac ax +1 \right) \left(1+\frac 1{(-1)^nx^{n+1}}\right)} dx \\ & = \int_0^\infty \frac {(1+x)(-1)^nx^{n+1}}{\sqrt x \left(x^2+ax+1\right)\left(1+(-1)^nx^{n+1}\right)} dx & ...(2) \end{aligned}

From ( ( 1 ) + ( 2 ) ) / 2 ((1)+(2))/2 :

I ( a ) = 1 2 ∫ 0 ∞ 1 + x x ( x 2 + a x + 1 ) d x Let u 2 = x ⟹ 2 u d u = d x = ∫ 0 ∞ u 2 + 1 u 4 + a u 2 + 1 d u = ∫ 0 ∞ u 2 + 1 ( u 2 + a − a 2 − 4 2 ) ( u 2 + a + a 2 − 4 2 ) d u = ∫ 0 ∞ u 2 + 1 ( u 2 + α ) ( u 2 + β ) d u By partial fractions = 1 β − α ∫ 0 ∞ ( 1 − α u 2 + α − 1 − β u 2 + β ) d u = π 2 ( β − α ) ( 1 − α α − 1 − β β ) Note that β − α = a 2 − 4 = π 2 a 2 − 4 ( β − α β − α + α β α β ) and α β = 1 = π a 2 − 4 ( β − α ) = π a 2 − 4 ( β − α ) 2 = π a 2 − 4 β − 2 α β + α = π a − 2 a 2 − 4 = π a + 2 \begin{aligned} I(a) & = \frac 12 \int_0^\infty \frac {1+x}{\sqrt x \left(x^2+ax+1\right)} dx & \small \color{#3D99F6} \text{Let }u^2 = x \implies 2u \ du = dx \\ & = \int_0^\infty \frac {u^2+1}{u^4+au^2+1} \ du \\ & = \int_0^\infty \frac {u^2+1}{\left(u^2+{\color{#3D99F6}\frac {a-\sqrt{a^2-4}}2}\right)\left(u^2+{\color{#D61F06}\frac {a+\sqrt{a^2-4}}2}\right)} \ du \\ & = \int_0^\infty \frac {u^2+1}{\left(u^2+{\color{#3D99F6}\alpha}\right)\left(u^2+{\color{#D61F06}\beta}\right)} \ du & \small \color{#3D99F6} \text{By partial fractions} \\ & = \frac 1{\beta - \alpha} \int_0^\infty \left(\frac {1-\alpha}{u^2+\alpha} - \frac {1-\beta}{u^2+\beta}\right) du \\ & = \frac {\pi}{2(\beta - \alpha)} \left(\frac {1-\alpha}{\sqrt \alpha} - \frac {1-\beta}{\sqrt \beta}\right) & \small \color{#3D99F6} \text{Note that } \beta - \alpha = \sqrt{a^2-4} \\ & = \frac {\pi}{2\sqrt{a^2-4}} \left(\frac { \sqrt \beta-\alpha \sqrt \beta - \sqrt \alpha +\sqrt \alpha \beta}{\sqrt {\alpha \beta}} \right) & \small \color{#3D99F6} \text{and } \alpha \beta = 1 \\ & = \frac {\pi}{\sqrt{a^2-4}} \left(\sqrt \beta- \sqrt \alpha \right) \\ & = \frac {\pi}{\sqrt{a^2-4}} \sqrt{\left(\sqrt \beta- \sqrt \alpha \right)^2} \\ & = \frac {\pi}{\sqrt{a^2-4}} \sqrt{\beta- 2\sqrt {\alpha \beta} + \alpha} \\ & = \frac {\pi \sqrt{a-2}}{\sqrt{a^2-4}} \\ & = \frac \pi{\sqrt{a+2}} \end{aligned}

⟹ I ( 1 + 2 2 ) = π 1 + 2 2 + 2 = π 3 + 2 2 = π 1 + 2 = π ( 2 − 1 ) \implies I\left(1+2\sqrt 2\right) = \dfrac \pi{\sqrt{1+2\sqrt 2+2}} = \dfrac \pi{\sqrt{3+2\sqrt 2}} = \dfrac \pi{1+\sqrt 2} = \pi\left(\sqrt 2-1\right)

⟹ ( a + 1 ) ( b + 3 ) ( c + 5 ) = ( 1 + 1 ) ( 2 + 3 ) ( 1 + 5 ) = 60 \implies (a+1)(b+3)(c+5) = (1+1)(2+3)(1+5) = \boxed{60}

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