关于分段估计思想的专题讨论

$f命题：$设$fleft( x ight) in Cleft[ {0,1} ight]$，证明：$lim limits_{h o egin{array}{*{20}{c}}{{0^ + }} end{array}} int_0^1 {frac{h}{{{x^2} + {h^2}}}} fleft( x ight)dx = frac{pi }{2}fleft( 0 ight)$

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$f命题：$设$forall [alpha ,eta ] subset left[ {0, + infty } ight),fleft( x ight) in R[alpha ,eta ]$，且$int_0^{ + infty } {fleft( x ight)dx} $收敛，则对于常数$a>1$成立[mathop {lim }limits_{y o egin{array}{*{20}{c}}{{0^ + }} end{array}} int_0^{ + infty } {{a^{ - xy}}fleft( x ight)dx} = int_0^{ + infty } {fleft( x ight)dx} ]

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$f命题：$设$forall a in left( {0, + infty } ight),fleft( x ight) in Rleft[ {0,a} ight]$，且$mathop {lim }limits_{x o  + infty } fleft( x ight) = alpha $，则

[mathop {lim }limits_{t o egin{array}{*{20}{c}}
{{0^ + }}
end{array}} tint_0^{ + infty } {{e^{ - tx}}fleft( x ight)dx} = alpha ]

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$f命题：$设$forall left[ {a,b} ight] subset left( { - infty , + infty } ight),fleft( x ight) in Rleft[ {a,b} ight]$，且$int_{ - infty }^{ + infty } {left| {fleft( x ight)} ight|dx} $，则$Fleft( x ight) = int_{ - infty }^{ + infty } {fleft( x ight)cos left( {2pi tx} ight)dt} $在$left( { - infty , + infty } ight)$上一致连续

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$f命题：$设$fleft( x ight) in Cleft[ {0, + infty } ight)$，且$int_0^{ + infty } {varphi left( x ight)dx} $绝对收敛，则[mathop {lim }limits_{n o infty } int_0^{sqrt n } {fleft( {frac{x}{n}} ight)varphi left( x ight)dx}  = fleft( 0 ight)int_0^{ + infty } {varphi left( x ight)dx} ]

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$f命题：$

附录

$f（Tauber定理）$设幂级数$sumlimits_{n = 0}^infty  {{a_n}{x^n}}  = Sleft( x ight)$在$left( { - 1,1} ight)$上成立，若$lim limits_{x o egin{array}{*{20}{c}}{{1^ - }}end{array}} Sleft( x ight) = S$，且$lim limits_{n o infty } n{a_n} = 0$，则$sumlimits_{n = 0}^infty  {{a_n}}  = S$

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$f（Riemman积分控制收敛定理）$设函数列$left{ {{f_n}left( x ight)} ight}$在$left[ {a, + infty } ight)$上内闭可积且一致收敛于$f(x)$，若存在函数$g(x)$，使得$left| {{f_n}left( x ight)} ight| le gleft( x ight)$对每个$x$与$n$都成立，且$int_a^{ + infty } {gleft( x ight)dx} $收敛，则[mathop {lim }limits_{n o infty } int_a^{ + infty } {{f_n}left( x ight)dx} { m{ = }}int_a^{ + infty } {fleft( x ight)dx} ]