拉普拉斯反变换 [p*(e^-p)]/1+(p^2) 同时另外一个问题 :Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int
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