设广义积分∫(e→+无穷)f(x)dx收敛,且满足方程f(x)=2/(除以)x^2-1/(除以)x乘以lnx的平方 ∫(e→+无
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![设广义积分∫(e→+无穷)f(x)dx收敛,且满足方程f(x)=2/(除以)x^2-1/(除以)x乘以lnx的平方 ∫(e→+无](/uploads/image/z/5937563-11-3.jpg?t=%E8%AE%BE%E5%B9%BF%E4%B9%89%E7%A7%AF%E5%88%86%E2%88%AB%EF%BC%88e%E2%86%92%2B%E6%97%A0%E7%A9%B7%EF%BC%89f%28x%29dx%E6%94%B6%E6%95%9B%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E6%96%B9%E7%A8%8Bf%28x%29%3D2%2F%EF%BC%88%E9%99%A4%E4%BB%A5%EF%BC%89x%5E2-1%2F%EF%BC%88%E9%99%A4%E4%BB%A5%EF%BC%89x%E4%B9%98%E4%BB%A5lnx%E7%9A%84%E5%B9%B3%E6%96%B9+%E2%88%AB%EF%BC%88e%E2%86%92%2B%E6%97%A0)
设广义积分∫(e→+无穷)f(x)dx收敛,且满足方程f(x)=2/(除以)x^2-1/(除以)x乘以lnx的平方 ∫(e→+无
设广义积分∫(e→+无穷)f(x)dx收敛,且满足方程f(x)=2/(除以)x^2-1/(除以)x乘以lnx的平方 ∫(e→+无
设广义积分∫(e→+无穷)f(x)dx收敛,且满足方程f(x)=2/(除以)x^2-1/(除以)x乘以lnx的平方 ∫(e→+无
Unexpectedly only me can help you?Don't mind I say English.
Let N = ∫(e→+∞) f(x) dx,since this integral is convergent,it's a constant
f(x) = 2/x² - 1/(xln²x) · ∫(e→+∞)
f(x) = 2/x² - 1/(xln²x) · N,integrate both sides with range from e to infinity
∫(e→+∞) f(x) dx = 2∫(e→+∞) 1/x² dx - N∫(e→+∞) 1/(xln²x) dx
N = 2 · - 1/x:(e→+∞) - N∫(e→+∞) 1/ln²x d(lnx)
N = - 2 · (0 - 1/e) - N · - 1/lnx:(e→+∞)
N = 2/e + N · (0 - 1)
N = 2/e - N
2N = 2/e
N = 1/e = ∫(e→+∞) f(x) dx
So f(x) = 2/x² - 1/(xln²x) · 1/e
f(x) = 2/x² - 1/(e · xln²x)