Tricky BEAST!!!!

Calculus Level 5

Evaluate

2 2010 ∫ 0 1 x 1004 ( 1 − x ) 1004 d x ∫ 0 1 x 1004 ( 1 − x 2010 ) 1004 d x . \frac {\displaystyle{ { 2 }^{ 2010 }\int _{ 0 }^{ 1 }{ { x }^{ 1004 } } ({ 1-x) }^{ 1004 }dx} }{\displaystyle{ \int _{ 0 }^{ 1 }{ { x }^{ 1004 } } { (1-{ x }^{ 2010 } })^{ 1004 }dx} }.

You may try its simpler version first WOWSOME


The answer is 4020.

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Kïñshük Sïñgh by Kïñshük Sïñgh , 28, India

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2 solutions

S o r r y f o r t h e i n c o n v e n i e n c e , i t s a n s w e r s h o u l d b e 4020. I ′ v e a n a n o t h e r s o l u t i o n f o r t h i s q u e s t i o n . S o l v i n g n u m e r a t o r : L e t x = s i n 2 θ d x = 2 s i n θ . c o s θ . d θ t h e r e f o r e , n u m e r a t o r t u r n s o u t t o b e : 2 2011 ∫ 0 π 2 s i n 2009 θ . c o s 2009 θ . d θ 4 ∫ 0 π 2 s i n 2009 2 θ . d θ . . . . . . . . . . . ( 1 ) N o w s o l v i n g d e n o m i n a t o r : L e t x 1005 = s i n θ 1005 x 1004 . d x = c o s θ . d θ x 1004 . d x = c o s θ . d θ 1005 t h e r e f o r e , d e n o m i n a t o r t u r n s o u t t o b e 1 1005 ∫ 0 π 2 c o s 2009 θ . d θ . . . . . . . . ( 2 ) N o w s u b s t i t u t i n g v a l u e s : 4020 ∫ 0 π 2 s i n 2009 2 θ . d θ ∫ 0 π 2 c o s 2009 θ . d θ = 4020 Sorry\quad for\quad the\quad inconvenience,\quad its\quad answer\quad should\quad be\quad 4020.\\ I've\quad an\quad another\quad solution\quad for\quad this\quad question.\\ Solving\quad numerator:\\ Let\quad x={ sin^{ 2 }\theta }\\ dx=2sin\theta .cos\theta .d\theta \\ therefore,\quad numerator\quad turns\quad out\quad to\quad be:\\ { 2 }^{ 2011 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin^{ 2009 } } \theta .{ cos }^{ 2009 }\theta .d\theta \\ 4\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2009 } } 2\theta .d\theta \quad ...........(1)\\ Now\quad solving\quad denominator:\\ Let\quad { x }^{ 1005 }=sin\theta \\ 1005{ x }^{ 1004 }.dx=cos\theta .d\theta \\ { x }^{ 1004 }.dx=\frac { cos\theta .d\theta }{ 1005 } \\ therefore,\quad denominator\quad turns\quad out\quad to\quad be\quad \\ \frac { 1 }{ 1005 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2009 } } \theta .d\theta \quad ........(2)\\ \\ Now\quad substituting\quad values:\\ \frac { 4020\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2009 } } 2\theta .d\theta }{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2009 } } \theta .d\theta } =\quad 4020\quad

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Mvs Saketh
Nov 9, 2014

My method might be long... i first converted lower integral into a form where gamma formula could be used.. then i applied gamma formula up and down and cancelled all terms and released the several 2s to cancel the already present 2^(2010) to leave 4.

Is there a shorter way?

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The problem asks us to compute 2 2010 ∫ 0 1 x 1004 ( 1 − x ) 1004 d x ∫ 0 1 x 1004 ( 1 − x 2010 ) 1004 d x . \frac{2^{2010} \int_0^1 x^{1004} (1 - x)^{1004} \: dx}{\int_0^1 x^{1004} (1 - x^{2010})^{1004} \: dx}. More generally, we compute 2 2 n ∫ 0 1 x n − 1 ( 1 − x ) n − 1 d x ∫ 0 1 x n − 1 ( 1 − x 2 n ) n − 1 d x , \frac{2^{2n} \int_0^1 x^{n - 1} (1 - x)^{n - 1} \: dx}{\int_0^1 x^{n - 1} (1 - x^{2n})^{n - 1} \: dx}, where n n is a positive integer.

By the Beta function , ∫ 0 1 x n − 1 ( 1 − x ) n − 1 d x = ( n − 1 ) ! ⋅ ( n − 1 ) ! ( 2 n − 1 ) ! . \int_0^1 x^{n - 1} (1 - x)^{n - 1} \: dx = \frac{(n - 1)! \cdot (n - 1)!}{(2n - 1)!}.

For the denominator, let y = x 2 n y = x^{2n} . Then x = y 1 / ( 2 n ) x = y^{1/(2n)} , and d x = 1 2 n y 1 / ( 2 n ) − 1 d y dx = \frac{1}{2n} y^{1/(2n) - 1} \: dy , so ∫ 0 1 x n − 1 ( 1 − x 2 n ) n − 1 d x = ∫ 0 1 y ( n − 1 ) / ( 2 n ) ( 1 − y ) n − 1 ⋅ 1 2 n y 1 / ( 2 n ) − 1 d y = 1 2 n ∫ 0 1 y − 1 / 2 ( 1 − y ) n − 1 d y . \begin{aligned} &\int_0^1 x^{n - 1} (1 - x^{2n})^{n - 1} \: dx \\ &= \int_0^1 y^{(n - 1)/(2n)} (1 - y)^{n - 1} \cdot \frac{1}{2n} y^{1/(2n) - 1} \: dy \\ &= \frac{1}{2n} \int_0^1 y^{-1/2} (1 - y)^{n - 1} \: dy. \end{aligned}

Again by the Beta function, this is equal to 1 2 n ⋅ Γ ( 1 2 ) ⋅ ( n − 1 ) ! Γ ( n + 1 2 ) . \frac{1}{2n} \cdot \frac{\Gamma (\frac{1}{2}) \cdot (n - 1)!}{\Gamma (n + \frac{1}{2})}. From the identity Γ ( t + 1 ) = t Γ ( t ) \Gamma(t + 1) = t \Gamma(t) , Γ ( 1 2 ) ⋅ ( n − 1 ) ! Γ ( n + 1 2 ) = ( n − 1 ) ! ⋅ Γ ( 1 2 ) Γ ( 3 2 ) ⋅ Γ ( 3 2 ) Γ ( 5 2 ) ⋯ Γ ( n − 1 2 ) Γ ( n + 1 2 ) = ( n − 1 ) ! ⋅ 2 1 ⋅ 2 3 ⋯ 2 2 n − 1 = ( n − 1 ) ! ⋅ 2 n 1 ⋅ 3 ⋅ 5 ⋯ ( 2 n − 1 ) = ( n − 1 ) ! ⋅ 2 n ⋅ 2 ⋅ 4 ⋯ ( 2 n ) 1 ⋅ 2 ⋯ ( 2 n ) = ( n − 1 ) ! ⋅ 2 n ⋅ 2 n ⋅ n ! ( 2 n ) ! . \begin{aligned} \frac{\Gamma (\frac{1}{2}) \cdot (n - 1)!}{\Gamma (n + \frac{1}{2})} &= (n - 1)! \cdot \frac{\Gamma(\frac{1}{2})}{\Gamma (\frac{3}{2})} \cdot \frac{\Gamma(\frac{3}{2})}{\Gamma (\frac{5}{2})} \dotsm \frac{\Gamma(n - \frac{1}{2})}{\Gamma (n + \frac{1}{2})} \\ &= (n - 1)!\cdot \frac{2}{1} \cdot \frac{2}{3} \dotsm \frac{2}{2n - 1} \\ &= (n - 1)! \cdot \frac{2^n}{1 \cdot 3 \cdot 5 \dotsm (2n - 1)} \\ &= (n - 1)! \cdot \frac{2^n \cdot 2 \cdot 4 \dotsm (2n)}{1 \cdot 2 \dotsm (2n)} \\ &= (n - 1)! \cdot \frac{2^n \cdot 2^n \cdot n!}{(2n)!}. \end{aligned}

Therefore, 2 2 n ∫ 0 1 x n − 1 ( 1 − x ) n − 1 d x ∫ 0 1 x n − 1 ( 1 − x 2 n ) n − 1 d x = 2 2 n ⋅ ( n − 1 ) ! ⋅ ( n − 1 ) ! ( 2 n − 1 ) ! 1 2 n ⋅ 2 2 n ⋅ ( n − 1 ) ! ⋅ n ! ( 2 n ) ! = 2 n ⋅ ( n − 1 ) ! ⋅ ( 2 n ) ! n ! ⋅ ( 2 n − 1 ) ! = 4 n . \begin{aligned} \frac{2^{2n} \int_0^1 x^{n - 1} (1 - x)^{n - 1} \: dx}{\int_0^1 x^{n - 1} (1 - x^{2n})^{n - 1} \: dx} &= \frac{\frac{2^{2n} \cdot (n - 1)! \cdot (n - 1)!}{(2n - 1)!}}{\frac{1}{2n} \cdot \frac{2^{2n} \cdot (n - 1)! \cdot n!}{(2n)!}} \\ &= \frac{2n \cdot (n - 1)! \cdot (2n)!}{n! \cdot (2n - 1)!} \\ &= 4n. \end{aligned}

For n = 1005 n = 1005 , this is 4 ⋅ 1005 = 4020 4 \cdot 1005 = 4020 .

Jon Haussmann - 6 years, 7 months ago

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Thanks. I have updated the answer accordingly.

Calvin Lin Staff - 6 years, 7 months ago

I don't think so there is any short method for this question. I too have solved this using gamma formula.

Kïñshük Sïñgh - 6 years, 7 months ago

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Can you review Jon's comment and let me know if anything is wrong with it?

Calvin Lin Staff - 6 years, 7 months ago

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