# How to bound this integral?

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.

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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
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

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.)

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D'oh, didn't think simple enough (+1). Out of curiosity, why did you use $\frac{1}{w}\log \frac{1}{1-w}$? I don't see how the $\frac{1}{w}$ enters the game here? – Daniel Fischer Apr 17 '14 at 12:51
@DanielFischer, when $w=e^{-2\pi y}$, the $dy$ becomes a constant times $dw/w$. – Barry Cipra Apr 17 '14 at 13:12
Ah, I didn't make the substitution, I let $\int e^{-2\pi ky}\,dy$ stand. – Daniel Fischer Apr 17 '14 at 13:16
@DanielFischer, ah yes, I guess I included an unnecessary substitition. – Barry Cipra Apr 17 '14 at 13:23
I see. I did compute the second integral via mathematica and it give me $\frac{1}{12}$, but I do not know how to use it properly. Thanks! – Bombyx mori Apr 17 '14 at 19:58