With the help of Hermite's Integral,I got
∞∑n=11n∫∞2πnsinxxdx=π−π2ln(2π)
I'd like to know can we solve this one using fourier series?
Answer
We begin with the expression
S=∞∑n=11n∫∞2πnsin(x)xdx
Enforcing the substitution x→2πnx in (1) yields
∞∑n=11n∫∞2πnsin(x)xdx=∞∑n=11n∫∞1sin(2πnx)xdx=∫∞11x∞∑n=1sin(2πnx)ndx=∞∑k=1∫k+1k1x(π2(1−2(x−k)))dx=π2∞∑k=1((2k+1)log(k+1k)−2)
where we recognized the Fourier sine series for π2(1−2x) for x∈(0,1) to go from (2) to (3).
To evaluate the sum in (4), we write the partial sum SK
SK=K∑k=1((2k+1)log(k+1k)−2)=−2K+K∑k=1((2k+1)log(n+1)−(2k−1)log(n))−2K∑k=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)+2−2log(√2πK(Ke)K)+O(1K)=2−log(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=11n∫∞2πnsin(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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