∫ 0 π / 2 ⎝ ⎛ n = 1 ∑ 5 sin n x ⎠ ⎞ ⎝ ⎛ m = 1 ∑ 5 cos m x ⎠ ⎞ d x
The closed form of the above integral can be represented as b a + d c π , where ( a , b ) and ( c , d ) are pairwise coprime positive integers.
Evaluate a + b + c + d .
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Under what conditions can we change the order of integration and summations?
P.S. In the fifth line from the top you've mistakenly written
where Γ ( x ) is the Gamma Beta function
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There are only finite terms added together, therefore integration and summations are interchangeable. Thanks for pointing out the typo.
Any short method ?? Although very less probability
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I = ∫ 0 2 π n = 1 ∑ 5 sin n x m = 1 ∑ 5 cos m x d x = ∫ 0 2 π n = 1 ∑ 5 m = 1 ∑ 5 sin n x cos m x d x = n = 1 ∑ 5 m = 1 ∑ 5 ∫ 0 2 π sin n x cos m x d x = 2 1 n = 1 ∑ 5 m = 1 ∑ 5 B ( 2 n + 1 , 2 m + 1 ) = 2 1 n = 1 ∑ 5 m = 1 ∑ 5 Γ ( 2 n + m + 1 ) Γ ( 2 n + 1 ) Γ ( 2 m + 1 ) = 2 1 ( 3 0 9 1 + 6 3 2 0 8 + 1 2 8 3 5 π ) = 1 2 6 0 3 9 9 1 + 2 5 6 3 5 π where B ( m , n ) is beta function. where Γ ( x ) is gamma function. See Note.
⟹ a + b + c + d = 3 9 9 1 + 1 2 6 0 + 3 5 + 2 5 6 = 5 5 4 2
Note:
When both m and n are odd:
S o o = B ( 1 , 1 ) + 2 B ( 1 , 2 ) + 2 B ( 1 , 3 ) + B ( 2 , 2 ) + 2 B ( 2 , 3 ) + 2 B ( 3 , 3 ) = 1 ! 0 ! 0 ! + 2 × 2 ! 0 ! 1 ! + 2 × 3 ! 0 ! 2 ! + 3 ! 1 ! 1 ! + 2 × 4 ! 1 ! 2 ! + 5 ! 2 ! 2 ! = 1 + 1 + 3 2 + 6 1 + 6 1 + 3 0 1 = 3 0 9 1
When either m or n is odd and the other even:
S o e = 2 [ B ( 1 , 2 3 ) + B ( 1 , 2 5 ) + B ( 2 , 2 3 ) + B ( 2 , 2 5 ) + B ( 3 , 2 3 ) + B ( 3 , 2 5 ) ] = 2 [ 2 3 ⋅ 2 1 π 0 ! ⋅ 2 1 π + 2 5 ⋅ 2 3 ⋅ 2 1 π 0 ! ⋅ 2 3 ⋅ 2 1 π + 2 5 ⋅ 2 3 ⋅ 2 1 π 1 ! ⋅ 2 1 π + 2 7 ⋅ 2 5 ⋅ 2 3 ⋅ 2 1 π 1 ! ⋅ 2 3 ⋅ 2 1 π + 2 7 ⋅ 2 5 ⋅ 2 3 ⋅ 2 1 π 2 ! ⋅ 2 1 π + 2 9 ⋅ 2 7 ⋅ 2 5 ⋅ 2 3 ⋅ 2 1 π 2 ! ⋅ 2 3 ⋅ 2 1 π ] = 2 [ 3 2 + 4 2 + 1 5 4 + 3 5 4 + 1 0 5 1 6 + 3 1 5 1 6 ] = 6 3 2 0 8
When both m and n are even:
S e e = B ( 2 3 , 2 3 ) + 2 B ( 2 3 , 2 5 ) + B ( 2 5 , 2 5 ) = 2 ! 2 1 π ⋅ 2 1 π + 2 × 3 ! 2 1 π ⋅ 2 3 ⋅ 2 1 π + 4 ! 2 3 ⋅ 2 1 π ⋅ 2 3 ⋅ 2 1 π = 8 1 π + 8 1 π + 1 2 8 3 π = 1 2 8 3 5 π