工程數學 積分問題

S(t^4 * e^-2t)dt=(-1/2)( S(t^4 * e^-2t(-2))dt
=(-1/2)( t^4 * e^-2t - S e^-2t*(4t^3)dt)…部分積分
=(-1/2)( t^4 * e^-2t) -(-1/2)4(-1/2) ( S e^-2t(-2)*(t^3)dt
第2次部分積分……第3次………
=(-1/2)( t^4 * e^-2t)- (-1/2)4(-1/2) ( t^3 * e^-2t)
-(-1/2)4(-1/2)3(-1/2)( ( t^2 * e^-2t)
-(-1/2)4(-1/2)3(-1/2) 2(-1/2) ( ( t * e^-2t)
-(-1/2)4(-1/2)3(-1/2) 2(-1/2) (-1/2) ( e^-2t)
…….上限t -下限t=0
得(!)=
(-1/2)( t^4 * e^-2t)- (-1/2)4(-1/2) ( t^3 * e^-2t)
-(-1/2)4(-1/2)3(-1/2)( ( t^2 * e^-2t)
-(-1/2)4(-1/2)3(-1/2) 2(-1/2) ( ( t * e^-2t)
-(-1/2)4(-1/2)3(-1/2) 2(-1/2) (-1/2) ( e^-2t) - 3/4
令 t 趨近於 無窮大
則 (!)趨近於 - 3/4

2010-01-09 20:59:37 補充：
謝謝天助大大指正!
訂正
=(-1/2)( t^4 * e^-2t)- (-1/2)4(-1/2) ( t^3 * e^-2t)

+(-1/2)4(-1/2)3(-1/2)( ( t^2 * e^-2t)

-(-1/2)4(-1/2)3(-1/2) 2(-1/2) ( ( t * e^-2t)

+(-1/2)4(-1/2)3(-1/2) 2(-1/2) (-1/2)
.........
(!)趨近於 3/4

Answer is 3/4 !