反常积分∫(0,∏)tandx= 求定积分∫(上π,下0)tanxdx
\u5b9a\u79ef\u5206(0-pie)tanxdx\u662f\u5426\u6536\u655b \u8981\u8be6\u89e3 \u6025\u5206\u4e3a0-pi/2\uff0cpi/2-pi\u4e0a\u7684\u5206\u6bb5\u53cd\u5e38\u79ef\u5206\uff0c\u90fd\u4e0d\u6536\u655b\uff0c\u56e0\u6b64\u539f\u5b9a\u79ef\u5206\u4e0d\u6536\u655b\u3002
\u56e0\u4e3a\u662f\u53cd\u5e38\u79ef\u5206\uff0c\u6240\u4ee5\u8981\u5206\u6210\uff1a
\u222b(\u4e0a\u03c0/2,\u4e0b0)tanxdx+\u222b(\u4e0a\u03c0,\u4e0b\u03c0/2)tanxdx
\u222b(\u4e0a\u03c0/2,\u4e0b0)tanxdx
=ln|secx|(\u4e0a\u03c0/2,\u4e0b0)
\u4e0a\u9650\u65f6\u7684\u503c\uff1a
lim(x->\u03c0/2-)ln|secx|
=\u65e0\u7a77\u5927
\u6240\u4ee5
\u672c\u5b9a\u79ef\u5206\u53d1\u6563\u3002
∫(π/2, 0)tanxdx+∫(π, π/2)tanxdx
∫(π/2, 0)tanxdx
=ln|secx|(π/2, 0)
上限时的值：
lim(x->π/2-)ln|secx|=∞
所以本定积分发散。
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