作业帮积分下限-∞,上限是+∞,求∫dx/[(x^2+1)^n]=?(n为自然数)
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积分下限-∞,上限是+∞,求∫dx/[(x^2+1)^n]=?(n为自然数)
积分下限-∞,上限是+∞,求∫dx/[(x^2+1)^n]=?(n为自然数)
积分下限-∞,上限是+∞,求∫dx/[(x^2+1)^n]=?(n为自然数)
设A(n)=(-∞,+∞)∫dx/[(x^2+1)^n],则
A(1)=(-∞,+∞)arctanx=π.
A(n)=(-∞,+∞)x/[(x^2+1)^n-) - (-∞,+∞)∫xd[1/(x^2+1)^n]（分布积分,前一项趋于0）
=-(-∞,+∞)∫xd[1/(x^2+1)^n]
=2n(-∞,+∞)∫x^2/[(x^2+1)^(n+1)]dx（将分子拆开：x^2=x^2+1-1,可得：）
=2nA(n)-2nA(n+1).
所以得递推公式：A(n+1)=[(2n-1)/(2n)]A(n) (n=1,2,...).因此：
A(n)=π[(1*3*5*...*(2n-3)]/[2*4*6*...*(2n-2)] (n=2,3,...).
§§§§§(n个)(-无穷,+无穷)d(arctanx)dydz...dn=§§§§pi
所以n=1时,积分为pi,n不等于1时，积分为+无穷
1
∫dx/(x^2+1)^n
x=tgu,x^2+1=(secu)^2 u=arctgx
∫dx/(x^2+1)^n=∫dtgu/(secu)^(2n)
∫dx/(x^2+1)=∫dtgu/secu^2=tgu/secu^2-∫tgudsecu/secu^3=tgu/secu^2-∫tgu*tgu*secudu/secu^3
...
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1
∫dx/(x^2+1)^n
x=tgu,x^2+1=(secu)^2 u=arctgx
∫dx/(x^2+1)^n=∫dtgu/(secu)^(2n)
∫dx/(x^2+1)=∫dtgu/secu^2=tgu/secu^2-∫tgudsecu/secu^3=tgu/secu^2-∫tgu*tgu*secudu/secu^3
=tgu/secu^2-∫(tgu)^2du/secu^2=tgu/secu^2-2∫dtgu/secu^2+2∫du/secu^2
∫dtgu/secu^2=∫du=u+C
∫dtgu/secu^(2n)=tgu/secu^(2n)-(2n)∫tgudsecu/secu^(2n)=tgu/secu^(2n)-(2n)∫dtgu/secu^(2n)+2n∫du/secu^(2n)
∫du/secu^(2n)=[-tgu/secu^(2n)+(2n+1)∫dtgu/secu^(2n)]/2n
∫du/secu^(2n-2)=[-tgu/secu^(2n-2)+(2n-1)∫dtgu/secu^(2n-2)]/(2n-2)
=[-tgu/secu^(2n-2)+(2n-1)∫du/secu^(2n-4)]/(2n-2)
2
tgu/secu^(2n-2)=x/(1+x^2)^(n-1), x→∞, x/(1+x^2)^(n-1)→0
x→-∞,x/(1+x^2)^(n-1)→0
∫[-∞∞] dx/(1+x^2)=∫[-π/2, π/2]du=π
∫[-∞,∞]dx/(1+x^2)^2=∫[-π/2,π/2]dtgu/secu^4=∫[-π/2,π/2]du/secu^2=3π/2
∫[-∞,∞]dx/(1+x^2)^3=∫[-π/2,π/2]dtgu/secu^6=∫[-π/2,π/2]du/secu^4=5*(3π/2)/4
∫[-,∞]dx/(1+x^2)^4=7*(5)*(3π/2)/4/8
∫[-,∞]dx/(1+x^2)^n=π*[(3*5*..*(2n-1)]/[2*4*..*(2n-2)] (n>1)
∫[-,∞]dx/(1+x^2)=π
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