各位版大好，請問以下積分的作法：
（不好意思我太會弄格式）
integrage θ: [0~2Pi]
integrand = Cos[nθ] /( 1 + 2ACos[θ] + A^2)
-1 < A < 1, n = 0, 1, 2, 3, ....
我現在看的課本用一個很間接的方式來求，我有看懂，可是我印象中看過其他比較好
的作法，請各位協助，謝謝～
p.s.
我的課本考慮 z == Exp[iθ], and -1 < t < 1
integrate along unit circle: 1 / {(z-A)(1/z - A )(1-tz)}
then by Residue theorem one easily obtains the integration result,
call it f(A, t).
On the other hand, 1/(1-tz) can be expanded as:
1 + tz + (tz)^2 + (tz)^3 +...
hence the similar expansion of f(A, t) yields an infinite series in t as well,
and the real part of the coefficient of t^n is the desired answer.
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