Question

e^(-x^2) 从0到inf 的定积分

Answer 1

先求不定积分：
∫
e^[(1+y)t]
*
sint
dt
z
=
1+y，只是简化常数项，不包括自变数t
=
∫
e^zt
*
sint
dt
=
-∫
e^zt
dcost
=
-e^zt
*
cost
+
z*∫
cost
*
e^zt
dt，分部积分法
=
-e^zt
*
cost
+
z*∫
e^zt
dsint
=
-e^zt
*
cost
+
z*e^zt
*
sint
-
z²*∫
sint
*
e^zt
dt，分部积分法，后移项
(1+z²)∫
e^zt
*
sint
dt
=
z*e^zt
*
sint
-
e^zt
*
cost
=
e^zt
*
(z*sint
-
cost)
∫
e^zt
*
sint
dt
=
e^zt
*
(z*sint
-
cost)
/
(1+z²)
+
c，之后代回常数项变换
∫
e^[(1+y)t]
*
sint
dt
=
e^[(1+y)t]
*
[(1+y)sint
-
cost]
/
[1+(1+y)²]
+
c'
定积分：将定积分∫(a->b)
f(t)
dt
=
lim(t->b)
F(t)
-
lim(t->a)
F(t)化为极限计算
∫(0->inf)
e^[(1+y)t]
*
sint
dt
=
1/[1+(1+y)²]
*
{lim(t->inf)
e^[(1+y)t]
*
[(1+y)sint
-
cost]
-
lim(t->0)e^[(1+y)t]
*
[(1+y)sint
-
cost]}
=
1/[1+(1+y)²]
*
[0
-
(-1)]
=
1/[1+(1+y)²]