Question Number 149773 by Samimsultani last updated on 07/Aug/21

Answered by puissant last updated on 07/Aug/21

I=∫_0 ^∞ ((cos(ax))/(x^2 +b^2 ))dx =(1/2)∫_(−∞) ^(+∞) ((cos(ax))/(x^2 +b^2 ))dx car cos(x) est paire =(1/2)Re(∫_(−∞) ^(+∞) (e^(aix) /(x^2 +b^2 ))dx) or ψ=∫_(−∞) ^(+∞) (e^(iax) /(x^2 +b^2 ))dx = 2iπ Res(ψ,ib) (theoreme du residu.).. =2iπ(e^(ia(ib)) /(2ib)) =(π/b)e^(−ab) d′ou ∵ I=(π/(2b))e^(−ab) .... ....Le puissant....