mma 给出太复杂的积分结果

hbghlyj

$a\inR,$$$I(a)=\int_{-\infty }^{\infty } \frac{\cos (a x)}{x^2+2 x+2} \, dx$$

1. Integrate[Cos[a x]/(x^2 + 2 x + 2), {x, -Infinity, Infinity}, Assumptions -> a \[Element] Reals]

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Mathematica

1. (1/(4 a))E^((-1-I) a) \[Pi] (-((1+E^(2 I a)) (-1+E^(2 a)) Abs[a])+a (-1+E^(2 I a)+E^(2 a)-E^((2+2 I) a)-2 (1+E^((2+2 I) a)) Ceiling[(\[Pi]-4 Arg[a])/(8 \[Pi])] Floor[(\[Pi]+Arg[a])/(2 \[Pi])]-2 Floor[(\[Pi]-4 Arg[a])/(8 \[Pi])] Floor[(\[Pi]+Arg[a])/(2 \[Pi])]-2 E^((2+2 I) a) Floor[(\[Pi]-4 Arg[a])/(8 \[Pi])] Floor[(\[Pi]+Arg[a])/(2 \[Pi])]+2 E^(2 I a) Floor[3/8-Arg[a]/(2 \[Pi])]+2 E^(2 a) Floor[3/8-Arg[a]/(2 \[Pi])]-2 E^(2 I a) Floor[(\[Pi]+Arg[a])/(2 \[Pi])] Floor[3/8-Arg[a]/(2 \[Pi])]-2 E^(2 a) Floor[(\[Pi]+Arg[a])/(2 \[Pi])] Floor[3/8-Arg[a]/(2 \[Pi])]+2 Ceiling[Arg[a]/(2 \[Pi])] ((-1+E^(2 I a)) (-1+E^(2 a))-(E^(2 I a)+E^(2 a)) Floor[3/8-Arg[a]/(2 \[Pi])])))

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hbghlyj

For $a>0$, by Jordan's lemma, the integral $\int{e^{iaz}\over z^2+2z+2}$ on semicircle contour $\to0$.

1. Re[2Pi I Residue[Exp[I a x]/(x^2+2x+2),{x,-1+I}]]//ComplexExpand

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$I(a)=\pi e^{-a}\cos (a)$
So, for $a\inR$, $I(a)=\pi e^{-\abs a}\cos (a)$

TSC999

用你的代码在我的 9.0 版本 MMA 上运行，结果是不用留数也行：