题目
已知 ∫ 0 + ∞ sin x x d x = π 2 \int_0^{ + \infty } {{{\sin x} \over x}dx} = {\pi \over 2} ∫0+∞xsinxdx=2π,求 ∫ 0 + ∞ sin 2 x x 2 d x \int_0^{ + \infty } {{{{{\sin }^2}x} \over {{x^2}}}dx} ∫0+∞x2sin2xdx.
来源
证明
∫
0
+
∞
sin
2
x
x
2
d
x
=
−
sin
2
x
x
∣
0
+
∞
+
∫
0
+
∞
sin
2
x
x
d
x
=
2
x
=
t
∫
0
+
∞
sin
t
t
d
t
=
π
2
.
\int_0^{ + \infty } {{{{{\sin }^2}x} \over {{x^2}}}dx} = \left. { - {{{{\sin }^2}x} \over x}} \right|_0^{ + \infty } + \int_0^{ + \infty } {{{\sin 2x} \over x}dx} \mathop = \limits^{2x = t} \int_0^{ + \infty } {{{\sin t} \over t}dt} = {\pi \over 2}.
∫0+∞x2sin2xdx=−xsin2x
0+∞+∫0+∞xsin2xdx=2x=t∫0+∞tsintdt=2π.
其中
lim
x
→
0
+
sin
2
x
x
=
lim
x
→
+
∞
sin
2
x
x
=
0.
\mathop {\lim }\limits_{x \to {0^ + }} {{{{\sin }^2}x} \over x} = \mathop {\lim }\limits_{x \to + \infty } {{{{\sin }^2}x} \over x} = 0.
x→0+limxsin2x=x→+∞limxsin2x=0.