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The question is as follows:

Find the integral $\int_0^1 \int_0^1 \sqrt{\frac{ts}{t^2 + s^2}} \,dt \,ds$.

$\textbf{Some effort:}$

We will use the Polar coordinates. Let $t = r \sin \theta$ and $s = r \cos \theta $. Then $dt ds = r \,dr \,d \theta$. I also need to show that it is finite.

The important part is to find the range of integration. I do not know how to find the bounds here? Maybe because our previous range is $1 \times 1$ square in the origion, so our $\theta$ will range from $0$ to $\frac{\pi}{2}$ and $r$ will range from 0 to 1?

But after changing the coordinates, we have $\int \int \sqrt{\frac{ts}{t^2 + s^2}} \,dt \,ds = \int \int r \sqrt{\sin \theta \cos \theta } \,dr \,d\theta$.

But here also there is an another problem that this integration has no elementary function linked to this result.

So can you please help me to show that $$\int_0^1 \int_0^1 \sqrt{\frac{ts}{t^2 + s^2}} \,dt \,ds < +\infty \text{?}$$ Thanks!

Harry Peter
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user510716
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  • @M.Strochyk Thanks! Yes we have it in David Bowman's answer. – user510716 Dec 13 '17 at 07:03
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    I mean that simple estimate is $| \sin{\theta} \cos{\theta} | = \frac{1}{2} | \sin{2\theta}| \leqslant\frac{\sqrt{2}}{4}$ for $0 \leqslant \theta \leqslant \frac{\pi}{2}.$ – M. Strochyk Dec 13 '17 at 07:06
  • @M.Strochyk Thanks! But you know, I do not know the bounds for my Polar coordinate system. What I wrote is not exact. Do you know what will be my bounds and why? – user510716 Dec 13 '17 at 07:27

2 Answers2

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$$(t-s)^2 = t^2 +s^2 -2ts \ge 0 \Rightarrow ts \le \frac{t^2+s^2}{2}.$$

So your integrand is bounded like $$\sqrt{\frac{ts}{t^2+s^2}}\le \sqrt{\frac{t^2+s^2}{2(t^2+s^2)}} = \frac{\sqrt{2}}{2}.$$

Integrating this over a compact domain gives a finite answer.

David Bowman
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$$\begin{eqnarray*}\iint_{(0,1)^2}\sqrt{\frac{xy}{x^2+y^2}}\,dx\,dy&\stackrel{\text{symmetry}}{=}&2\int_{0}^{1}\int_{0}^{x}\sqrt{\frac{xy}{x^2+y^2}}\,dy\,dx\\&\stackrel{y\mapsto x u}{=}&2\int_{0}^{1}\int_{0}^{1}x \sqrt{\frac{x^2 u}{x^2+x^2 u^2}}\,du\,dx\\&\stackrel{\text{Fubini}}{=}&2\int_{0}^{1}x\,dx \int_{0}^{1}\sqrt{\frac{u}{1+u^2}}\,du\\&=&\int_{0}^{1}\sqrt{\frac{u}{1+u^2}}\,du\\&\stackrel{u\mapsto v^2}{=}&\int_{0}^{1}\frac{2v^2}{\sqrt{1+v^4}}\,dv\leq\int_{0}^{1}2v^2\,dv=\frac{2}{3}.\end{eqnarray*}$$ Better bounds can be derived from the Cauchy-Schwarz or Jensen's inequalities.

Jack D'Aurizio
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