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If they are equal, sum them up!#ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

It's nice for x→1/x to get the first part.
But can't understand why the required integral [∫1 /(1+x^2) ^2dx] equal to 1/2 of ∫1/(1+x^2) dx?

This is trivial.#ed_op#br#ed_cl#
Make the substitution x -> 1/x we get the first part.#ed_op#br#ed_cl#
It suffices to consider int_0^infinity 1/(1+x^2) dx = arctanx | 0 -> infinity = pi/2#ed_op#br#ed_cl#
so required integral = 1/2 * pi/2 = pi/4#ed_op#br#ed_cl#
#ed_op#br#ed_cl#
Why should one use advanced method for a trivial question?#ed_op#br#ed_cl#

#ed_op#DIV#ed_cl#用到傅立葉正餘弦變換。小寫表示原函數，大寫表示經變換後的函數，#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#下標s,c表示正弦或餘弦(sin,cos)。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#式中亦用到逆變換形式。簡單來說，如果熟悉傅利葉正餘弦變換形式，#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#並且可以容易拆解為兩個同為正弦或餘弦變換，則可以用此方法計算為便捷。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這跟拉式變換一樣都是一種積分轉換，此法是由傅立葉變換拆解實部與虛部而來。#ed_op#/DIV#ed_cl#

算法精簡, 但是很難理解.
請問在原因中Fs(w), Gs(w), f(t), g(t)各代表什麼?

#ed_op#DIV#ed_cl#最快的方法，但是很難想到#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#P align=left#ed_cl##ed_op#IMG alt="image file name: 2k4c6f031f30.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k4c6f031f30.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#P align=center#ed_cl##ed_op#B#ed_cl##ed_op#FONT face=Verdana size=2#ed_cl##ed_op#A href="http://yll.loxa.edu.tw/phpBB2/richedit/show_it.php?step=upload"#ed_cl##ed_op#/A#ed_cl##ed_op#/FONT#ed_cl##ed_op#/B#ed_cl# #ed_op#/P#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

∫1 /(1+X^2) ^2dX at((0, ∞)
令X=tan Y, dX/dY=sec^2Y  ,, Y at (0, π/2)
∫1 /(1+X^2) ^2dX=∫1 /(1+tan^2Y) ^2 *sec^2Y dY =∫1 /sec^2Y dY=
∫cos^2Y dY at (0, π/2)
=∫(1+cos2Y)/2 dY=1/2∫(1+cos2Y) dY=1/2(Y| at (0, π/2)+ (1/2)sin2Y| at (0, π/2))
=π/4

令X=tan Y,
∫X^2 /(1+X^2) ^2dX=∫tan^2Y /(1+tan^2Y) ^2 *sec^2Y dY =
∫tan^2Y /sec^2Y dY=∫sin^2Y dY=∫cos^2Y dY at (0, π/2)

#ed_op#DIV#ed_cl#證明：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kde74e00300.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kde74e00300.png" border=0#ed_cl##ed_op#/DIV#ed_cl#