Question:1∫0dxloglog1x(1+x)2=12logπ2−γ2
I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation x↦−logx to get rid of the nested log. ThereforeI=∞∫0dxe−xlogx(1+e−x)2
The inside integrand can be rewritten as an infinite series to getI=∑n≥0(n+1)(−1)n∞∫0dxe−x(n+1)logxThe inside integral, I thought, could be evaluated by differentiating the gamma function to get∞∫0dte−t(n+1)logt=−γn+1−log(n+1)n+1
However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that∑n≥0(−1)n=12Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.
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