Integration (1)

Calculus Level 4

0 1 ( 1 t 50 ) 100 d t 0 1 ( 1 t 50 ) 101 d t \large \dfrac{\displaystyle\int_0^1 (1-t^{50})^{100} \, dt}{\displaystyle\int_0^1 (1-t^{50})^{101} \, dt}

If the value of the expression above is equal to p q \dfrac pq , where p p and q q are coprime positive integers, find p q p-q .


The answer is 1.

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1 solution

Chew-Seong Cheong
Jan 21, 2016

0 1 ( 1 t 50 ) 100 d x 0 1 ( 1 t 50 ) 101 d x = 1 50 0 1 x 49 50 ( 1 x ) 100 d x 1 50 0 1 x 49 50 ( 1 x ) 101 d x Let x = t 50 d x = 50 t 49 d t d t = x 49 50 d x 50 = 0 1 x 1 50 1 ( 1 x ) 101 1 d x 0 1 x 1 50 1 ( 1 x ) 102 1 d x = B ( 1 50 , 101 ) B ( 1 50 , 102 ) B ( x , y ) is Beta function = Γ ( 1 50 ) Γ ( 101 ) Γ ( 101 1 50 ) × Γ ( 102 1 50 ) Γ ( 1 50 ) Γ ( 102 ) Γ ( x ) is Gamma function = 101 1 50 101 = 5051 5050 \begin{aligned} \frac{\displaystyle \int_0^1 (1-t^{50})^{100} dx}{\displaystyle \int_0^1 (1-t^{50})^{101} dx} & = \frac{ \frac{1}{50} \displaystyle \int_0^1 x^{-\frac{49}{50}}(1-x)^{100} dx}{ \frac{1}{50} \displaystyle \int_0^1 x^{-\frac{49}{50}}(1-x)^{101} dx} \quad \quad \small \color{#3D99F6}{\text{Let }x = t^{50} \Rightarrow dx = 50 t^{49}dt \Rightarrow dt = \frac{x^{-\frac{49}{50}}dx}{50}} \\ & = \frac{\displaystyle \int_0^1 x^{\frac{1}{50}-1}(1-x)^{101-1} dx}{ \displaystyle \int_0^1 x^{\frac{1}{50}-1}(1-x)^{102-1} dx} \\ & = \frac{B\left(\frac{1}{50},101\right)}{B\left(\frac{1}{50},102\right)} \quad \quad \small \color{#3D99F6}{B(x,y) \text{ is Beta function}} \\ & = \frac{\Gamma\left(\frac{1}{50}\right)\Gamma\left(101\right)}{\Gamma\left(101\frac{1}{50}\right)} \times \frac{\Gamma\left(102\frac{1}{50}\right)}{\Gamma\left(\frac{1}{50}\right)\Gamma\left(102\right)} \quad \quad \small \color{#3D99F6}{\Gamma(x) \text{ is Gamma function}} \\ & = \frac{101\frac{1}{50}}{101} \\ & = \frac{5051}{5050} \end{aligned}

p q = 5051 5050 = 1 \Rightarrow p-q = 5051-5050 = \boxed{1}

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Aditya Kumar - 5 years, 4 months ago

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