幫補上第10題的過程
令 \( \displaystyle t = {x^2},dt = 2xdx \)，原式 \( \displaystyle = \int_0^\infty {\frac{1}{x} \times {e^{ - {x^2}}} \times 2xdx} = 2\int_0^\infty {{e^{ - {x^2}}}dx} \)
\( \displaystyle {\left( {\int_0^\infty {{e^{ - {x^2}}}dx} } \right)^2} = \int_0^\infty {{e^{ - {x^2}}}dx} \times \int_0^\infty {{e^{ - {y^2}}}dy} = \int_0^\infty {\int_0^\infty {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \)
\( \displaystyle = \int_0^{\frac{\pi }{2}} {\int_0^\infty {{e^{ - {r^2}}} \times rdrd\theta } } = \frac{\pi }{2}\int_0^\infty {r{e^{ - {r^2}}}dr} = \frac{\pi }{4} \)，故 \( \displaystyle \int_0^\infty {{e^{ - {x^2}}}dx} = \frac{{\sqrt \pi }}{2} \)
原式 \( \displaystyle = 2\int_0^\infty {{e^{ - {x^2}}}dx} = \sqrt \pi \)
然後想問一下laylay老師怎麼知道第9題要先拉出I而不是去找特徵根
[ 本帖最後由 BambooLotus 於 2017-6-12 00:40 編輯 ]
