Figure it out

Calculus Level 4

f ( x ) = ∫ 0 ∞ t ⋅ d t ( t + 1 ) x f\left( x \right) ={ \int _{ 0 }^{ \infty }{ \frac { t\cdot dt }{ { \left( t+1 \right) }^{ x } } } }

If f ( x ) f\left( x \right) is defined when x > 2 x>2 , which of the following is equivalent to f ( x ) f\left( x \right) ?

This problem is original.
Picture credits: Bi-crystal by Richard Bartz, Wikipedia
x x 2 + 2 x − 1 \frac { x }{ { x }^{ 2 }+2x-1 } x Γ ( x + 1 ) tan ⁡ − 1 x \frac { { x\Gamma (x+1) } }{ \tan ^{ -1 }{ x } } ln ⁡ x x x + 1 \frac { { \ln { x } } }{ { x }^{ x+1 } } ∞ \infty x 2 tan ⁡ − 1 x \frac { { x } }{ 2\tan ^{ -1 }{ x } } 1 x 2 + x − 4 \frac { 1 }{ { x }^{ 2 }+x-4 } 1 x 2 + 2 x − 1 \frac { 1 }{ { x }^{ 2 }+2x-1 } 1 x 2 − 3 x + 2 \frac { 1 }{ { x }^{ 2 }-3x+2 }

Jonas Katona by Jonas Katona , 22, USA

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1 solution

Hassan Abdulla
Sep 3, 2020

Recall Beta function : B ( a , b ) = ∫ 0 ∞ t a − 1 ( 1 + t ) a + b d t \color{#D61F06} B(a,b)=\int_0^\infty \frac{t^{a-1}}{(1+t)^{a+b}} \ dt

f ( x ) = B ( 2 , x − 2 ) = Γ ( 2 ) ⋅ Γ ( x − 2 ) Γ ( x ) = 1 ⋅ Γ ( x − 2 ) ( x − 1 ) ( x − 2 ) Γ ( x − 2 ) s i n c e Γ ( a + 1 ) = a ⋅ Γ ( a ) = 1 ( x − 1 ) ( x − 2 ) = 1 x 2 − 3 x + 2 \begin{aligned} f(x) &=B(2,x-2)\\ &=\frac{{\color{#20A900}\Gamma(2)} \cdot \Gamma(x-2)}{\color{#D61F06}\Gamma(x)} \\ &=\frac{{\color{#20A900}1} \cdot \Gamma(x-2)}{\color{#D61F06}(x-1)(x-2)\Gamma(x-2)} & & since \ \Gamma(a+1)=a \cdot \Gamma(a) \\ &=\frac{1}{(x-1)(x-2)} \\ &=\frac{1}{x^2-3x+2} \end{aligned}

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