作业帮证明∫sin(x^2)dx=0.5√(π/2),积分区间为0到正无穷.
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证明∫sin(x^2)dx=0.5√(π/2),积分区间为0到正无穷.
证明∫sin(x^2)dx=0.5√(π/2),积分区间为0到正无穷.
证明∫sin(x^2)dx=0.5√(π/2),积分区间为0到正无穷.
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∫sin(x^2)dx
=(1/2)∫(1/x)sin(x^2)d(x^2)
=-(1/2)∫(1/x)d[cos(x^2)]
设y=cos(x^2),
0≤x＜∞
-1≤y≤1
x=√arccosy
∫sin(x^2)dx=-(1/2)∫(1/x)d[cos(x^2)]
=-(1/2)∫(1/√arccosy)dy
f(t)=∫[0,+∞]sin(tx^2)dx
用拉普拉斯变换
F(s)=∫[0,+∞]f(t)e^(-st)dt
=∫[0,+∞]{∫[0,+∞]sin(tx^2)dx}e^(-st)dt
交换积分次序，先对t积分
F(s)=∫[0,+∞]dx∫[0,+∞]sin(tx^2)e^(-st)dt
=∫[0,+∞]x^2/(x^4+s^2)dx
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f(t)=∫[0,+∞]sin(tx^2)dx
用拉普拉斯变换
F(s)=∫[0,+∞]f(t)e^(-st)dt
=∫[0,+∞]{∫[0,+∞]sin(tx^2)dx}e^(-st)dt
交换积分次序，先对t积分
F(s)=∫[0,+∞]dx∫[0,+∞]sin(tx^2)e^(-st)dt
=∫[0,+∞]x^2/(x^4+s^2)dx
=π/(2√2)*1/√s
再取逆变换
1/√s的逆变换是1/√π*1/√t
故f(t)=√π/(2√2)*1/√t
t=1即为积分值
∫[0,+∞]sinx^2dx=f(1)=√(π/8)
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