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I want to ask how a hint how to show this integral inequality: $$ \frac{1}{\pi}\int^{\infty}_{0}\frac{x}{y^{2}+x^{2}}\log\frac{1}{1-e^{-2\pi y}}dy< \frac{1}{12x} $$ This is from Ahlfors, Complex Analysis, page $206$. I tried to compute a rough bound, but my bound is too rough and it did not work.

Explicit computation showed the bound is quite delicate, for example for $x=6000$ the value on the left hand side is $0.000013888888876028806686426283065381014133382406390603...$ as opposed to $0.00001388888888...$. For $x=60000$ the accuracy is about $0.99999999999074074074$, for $x=75000$ the accuracy is about $0.99999999999407407407$.

This is the refinement term in the usual Stirling's formula. I know it for a long time but I never knew how to prove it.

Bombyx mori
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  • It might not work but did you have a go at polar coordinates? – user88595 Apr 17 '14 at 10:43
  • The numeric integral value is wrong. – user2345215 Apr 17 '14 at 10:44
  • Up to 50 digits, it is:0.00013888888760288076050033800243611840892960169910069 – Bombyx mori Apr 17 '14 at 10:47
  • How is $0.0001\ldots$ less than $0.00001\ldots$ then? – user2345215 Apr 17 '14 at 10:48
  • It seems I had a typo in my command. The correct result should be 0.000013888888876028806686426283065381014133382406390603... – Bombyx mori Apr 17 '14 at 10:52
  • Anyway, thanks for pointing out! – Bombyx mori Apr 17 '14 at 10:53
  • @user88595: This is really a 1 variable integral. I do not think changing it into polar coordinates would simplify anything. – Bombyx mori Apr 17 '14 at 11:14
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    Dirty trick: instead of writing $\sum \frac{1}{(z+n)^2}$ as an integral of $\frac{\pi\cot \pi\zeta}{(z+\zeta)^2}$, use $\frac{1}{(z+n)^2} = \int_0^\infty te^{-(z+n)t},dt$. That leads to the expression $$\int_0^\infty e^{-zt}\left(\frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right)\frac{dt}{t}$$ for the refinement term, and $$0 \leqslant \frac{1}{t}\left(\frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right) \leqslant \frac{1}{12}$$ for $t\geqslant 0$ can be verified elementarily (but a little tediously). – Daniel Fischer Apr 17 '14 at 11:47
  • @DanielFischer: I see. This seems to be really helpful, since the integral is really difficult to evaluate. – Bombyx mori Apr 17 '14 at 11:58
  • It seems a bit like cheating, but looking at the other integral, I have no idea yet how to attack it directly. – Daniel Fischer Apr 17 '14 at 12:08

1 Answers1

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Hint 1:

$${x^2\over x^2+y^2}\lt1$$

Hint 2:

$${1\over w}\log{1\over1-w}=1+{1\over2}w+{1\over3}w^2+\cdots$$

(Note: I'll flesh this out if the OP requests, but all the questions asked for was a hint.)

Barry Cipra
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