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Question Number 150647 by puissant last updated on 14/Aug/21
Answered by puissant last updated on 14/Aug/21
posonsI2n=∫0π2(sint)2ndt=1×3×5......×(2n−1)2×4×6×.....×2n×π2⇒I2n=2n!22nn!×π2(integraledeWallis)oreixsint=∑n⩾0(eixsint)n!ona:∫−ππeixsintdt=∑n⩾0(ix)nn!∫−ππ(sint)ndt⇒∫−ππeixsintdt=∑n⩾0(−1)nx2n(2n)!(4∫0π2(sint)2ndt)=∑n⩾0(−1)nx2n(2n)!×4(2n)!22n(n!)2×π2=∑n⩾0(−1)nx2n22n(n!)2×2π⇒∫−ππeixsintdt=2πf(x)D′ouf(x)=12π∫−ππeixsintdt...............Lepuissant..........
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