Wednesday, 9 January 2013

calculus - Integral inti0nftyfraclog2xcosaxxn1dx

Hi I am trying to calculate
I:=0log2xcos(ax)xn1dx,(n)>1,aR.
Note if we set a=0 we get a similar integral given by
0log2xxn1dx=2π3cot(π/n)csc2(π/n)n3,(n)>1.
I was trying to write I as
I=[0eiaxlog2xxn1dx]=[0eiaxlog2x1xndx]=[0eiaxlog2xm=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.

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