Hi I am trying to calculate
I:=∞∫0log2xcos(ax)xn−1dx,ℜ(n)>1,a∈R.
Note if we set a=0 we get a similar integral given by
∞∫0log2xxn−1dx=−2π3cot(π/n)csc2(π/n)n3,ℜ(n)>1.
I was trying to write I as
I=ℜ[∞∫0eiaxlog2xxn−1dx]=ℜ[0∫∞eiaxlog2x1−xndx]=ℜ[0∫∞eiaxlog2x∞∑m=0xnmdx].
But was unsure of where to go from here. How can we calculate I? It is clear that this method is not going to work.
Wednesday, 9 January 2013
calculus - Integral inti0nftyfraclog2xcosaxxn−1dx
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