Thursday, 29 December 2016

calculus - How to evaluate this series using fourier series?



With the help of Hermite's Integral,I got
n=11n2πnsinxxdx=ππ2ln(2π)
I'd like to know can we solve this one using fourier series?


Answer




We begin with the expression



S=n=11n2πnsin(x)xdx



Enforcing the substitution x2πnx in (1) yields



n=11n2πnsin(x)xdx=n=11n1sin(2πnx)xdx=11xn=1sin(2πnx)ndx=k=1k+1k1x(π2(12(xk)))dx=π2k=1((2k+1)log(k+1k)2)



where we recognized the Fourier sine series for π2(12x) for x(0,1) to go from (2) to (3).



To evaluate the sum in (4), we write the partial sum SK



SK=Kk=1((2k+1)log(k+1k)2)=2K+Kk=1((2k+1)log(n+1)(2k1)log(n))2Kk=1log(k)=2K+(2K+1)log(K+1)2log(K!)=2K+(2K+1)log(K)+(2K+1)log(1+1K)2log(K!)=2K+(2K+1)log(K)+22log(2πK(Ke)K)+O(1K)=2log(2π)+O(1K)



In going from (5) to (6), we used the asymptotic expansion for the logarithm function, log(1+x)=x+O(x2) and Stirling's Formula



K!=(2πK(Ke)K)(1+O(1K))




Finally, putting it all together, we find that



S=n=11n2πnsin(x)xdx=π2lim



and therefore




\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n}\int_{2\pi n}^\infty \frac{\sin (x)}{x}\,dx=\pi-\frac{\pi}{2}\log(2\pi)}



as was to be shown!


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