Here, however, we can go a little further, since

$$ \frac{1}{m+n-1-\alpha} + \frac{1}{m+n-1+\alpha} > \frac{2}{m+n-1} > $$ for $ 0 < \alpha < 1 $, and sob

$$ \int^m_{m - 1}\int^n_{n - 1} \frac{dxdy}{x+y} > \frac{1}{m+n-1} $$

If now we replace $m$ and $n$ by $m+1$ and $n+1$, we obtain a slightly sharper form of Theorem 315, viz.

323. If the conditions of Theorem 315 are satisfied, then

$$ \sum^\infty_0 \sum^\infty_0 \frac{a_mb_n}{m + n + 1} < \frac{\pi}{sin(\pi/p)}\left( \sum^\infty_0 a^p_m\right)^{1/p} \left( \sum^\infty_0 b^{p'}_n\right)^{1/p'}. $$

a We describe Theorems 315 (together with the sharper Theorem 323) and 316 as 'Hilbert's theorems'. Strictly, Hilbert's Theorem is Theorem 315, with $p = 2$.

b Associate elements of the integral symmetrically situated about the centre of the square of integration.

(Screenshot for reference)

Now i can not understand $\int_{m-1}^m \int_{n-1}^n \frac{d x d y}{x+y}>\frac{1}{m+n-1}$. How to obtain $\int_{m-1}^m \int_{n-1}^n \frac{d x d y}{x+y}>\frac{1}{m+n-1}$? What variable substitution is needed to prove it?

remark:I just did some calculations, it seems that directly calculating the integral and then using the derivative to prove it should work, but the process is cumbersome. Is there any other good method? Or how should we interpret the footnote b in the book?


2 Answers 2


Note that, by letting $t=m-1/2$ and $s=n-1/2$, we get $$I:=\int_{m-1}^m \int_{n-1}^n \frac{d x d y}{x+y}= \int_{t=-1/2}^{1/2} \int_{s=-1/2}^{1/2} \frac{d s d t}{m+n-1+s+t}.$$ By replacing $t$ with $-t$ and $s$ with $-s$, we also have $$I=\int_{m-1}^m \int_{n-1}^n \frac{d x d y}{x+y}= \int_{t=-1/2}^{1/2} \int_{s=-1/2}^{1/2} \frac{d s d t}{m+n-1-s-t}.$$ Therefore $$I=\frac{1}{2}\int_{t=-1/2}^{1/2} \int_{s=-1/2}^{1/2}\left( \frac{1}{m+n-1+s+t} +\frac{1}{m+n-1-s-t}\right) dsdt$$ By AM-GM inequality, $1/a+1/b> 4/(a+b)$ when $a\not=b$ and $a,b>0$, therefore $$I=\frac{1}{2}\int_{t=-1/2}^{1/2} \int_{s=-1/2}^{1/2}\left( \frac{1}{m+n-1+s+t} +\frac{1}{m+n-1-s-t}\right) dsdt\\>\frac{1}{2}\int_{t=-1/2}^{1/2} \int_{s=-1/2}^{1/2}\frac{2}{m+n-1}dsdt=\frac{1}{m+n-1}.$$

  • $\begingroup$ sorry, Shouldn't the minimum value of \frac{1}{x+y} be \frac{1}{m+n}? $\endgroup$ Sep 23 at 8:58
  • $\begingroup$ @YilinCheng You are correct. Please see my edit. $\endgroup$
    – Robert Z
    Sep 23 at 10:06
  • 1
    $\begingroup$ Thank you very much, i understand your edit and now i understand the proof. $\endgroup$ Sep 23 at 10:25

As is mentioned in the text just above the result, for $\alpha \in (0,1)$, $$\frac{1}{m+n-1-\alpha}+\frac{1}{m+n-1+\alpha}>\frac{2}{m+n-1}$$ This follows from Jensen's inequality for $1/x$ or noting that $2x^2>2(x^2-\alpha^2)$ and hence $\frac{2x}{x^2-\alpha^2}>\frac{2}{x}$ giving that $\frac{1}{x-\alpha}+\frac{1}{x+\alpha}>\frac{2}{x}$. Now $\int_{m-1}^m\int_{n-1}^{n}\frac{1}{x+y} dx dy$ is the integral over a square so as footnote b mentions you can use the inequality to associate values symmetrically placed with respect to the square's centre to give the result.

However I think it may be easier just to use Jensen's inequality for the function $g(x,y)=\frac{1}{x+y}$ which is convex over the domain $(m-1,m)\times(n-1,n)$. Since $\int_{m-1}^m\int_{n-1}^{n}\frac{1}{x+y} dx dy$ is the mean value of $g$ over this domain it must be greater than value of $g$ at the domain centre, $g(\frac{m+n-2}{2},\frac{m+n}{2})=\frac{1}{m+n-1}.$

  • $\begingroup$ thanks you very much. $\endgroup$ Sep 23 at 10:55

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