[微積] 請教一題積分
※ 引述《JohnMash (Paul)》之銘言:
※ 引述《penny93730 (思考)》之銘言:
: ∞ sin x
: ∫ ------- dx
: 0 x
: 以上,感謝:)
Let me give another approach
2i∫[0,∞] sin x /x dx
= ∫[-∞,∞] (cos x + i sin x -1)/x dx
= ∫[-∞,∞] (e^{ix}-1)/x dx
= -lim_{R→∞}∫[0,π] (-1)/(R e^{iθ}) R i e^{iθ} dθ
= πi
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