Last Tribute to 2017
∫ 0 1 x 2 0 1 7 ( 1 − x ) 2 0 1 8 d x = γ ! α ! β !
The equation above holds true for integers α , β , and γ , where α < β . Submit α − β + γ − 1 .
The answer is 4034.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Direct consequence of Beta Function hence answer is simply β ( x , y ) = Γ ( x + y ) Γ ( x ) Γ ( y ) = Γ ( 4 0 3 7 ) Γ ( 2 0 1 8 ) Γ ( 2 0 1 9 ) = 4 0 3 6 ! 2 0 1 7 ! 2 0 1 8 ! ; α − β + γ − 1 = 2 0 1 7 − 2 0 1 8 + 4 0 3 6 − 1 = 4 0 3 4
Relevant wiki: Beta Function
I = ∫ 0 1 x 2 0 1 7 ( 1 − x ) 2 0 1 8 d x = B ( 2 0 1 8 , 2 0 1 9 ) = Γ ( 4 0 3 7 ) Γ ( 2 0 1 8 ) Γ ( 2 0 1 9 ) = 4 0 3 6 ! 2 0 1 7 ! 2 0 1 8 ! Since B ( m , n ) = ∫ 0 1 u m − 1 ( 1 − u ) n − 1 d u where B ( m , n ) is the beta function. Since B ( m , n ) = Γ ( m + n ) Γ ( m ) Γ ( n ) where gamma function Γ ( n ) = ( n − 1 ) !
Therefore, α − β + γ − 1 = 2 0 1 7 − 2 0 1 8 + 2 0 3 6 − 1 = 4 0 3 4 .