∫ 0 ∞ 1 + x 1 3 1 d x = b a π csc ( c π )
If the equation above holds true for positive integers a , b and c , with a , b coprime, find a + b + c .
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One can use contour Integration to show that ∫ x m / ( 1 + x n ) d x = ( π / n ) / s i n ( m + 1 ) π / n . with the given set of limits(zero-infinite).For us the integral is simply π / 1 3 c s c ( π / 1 3 )
A beautiful solution I have seen multiple times on Math Stack:
∫ 0 ∞ x n + 1 1 d x = ∫ 0 ∞ ∫ 0 ∞ e − t ( 1 + x n ) d x d t
Sub u = x n t d u = n x n − 1 t d x
n 1 ∫ 0 ∞ e − t t n − 1 d t ∫ 0 ∞ u n 1 − 1 e − u d u =
n Γ ( 1 − n 1 ) Γ ( n 1 ) = n sin n π π by Euler’s Reflection Formula
Plug in 13 for 1 3 sin 1 3 π π
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