I'm given the following improper integral: $\int_{D} \frac{dxdy}{\sqrt{1-a\cdot x-b\cdot y}}$ where $D$ is the open unit disk, assuming $a^2 + b^2 = 1$. I should decide whether it converges or not, and if it does, I should find the limit.

I've found that the integrand is well defined on $D$ (using Cauchy-Schwarz lemma) and by using this lemma I was able to upper-bound the integral by another and show that it converges. However, I cannot evaluate this integral, i.e. find its exact value. I'm quite sure I should integrate by substitution, but it is just not working... How can I do it?

Thanks in advance!

  • $\begingroup$ The disc is rotationally symmetric.... $\endgroup$ – Lord Shark the Unknown Jan 19 '18 at 17:32
  • $\begingroup$ @LordSharktheUnknown How can I use this fact? $\endgroup$ – RanSch Jan 19 '18 at 18:37

Hint 1:$\newcommand\dd{\mathrm d}\newcommand\INT{\int\limits}$

Find a linear transformation (substitution) such that the integral becomes $$\int_D \frac{\dd u\,\dd v}{\sqrt{1-u}}.$$

Hint 1 howto:

We have $$\INT_{x^2+y^2<1} \frac{\dd x\,\dd y}{\sqrt{1-ax-by}} = \INT_{u^2+v^2<1} \frac{\dd u\,\dd v}{\sqrt{1-u}}$$ with $u=ax+by$, $v=-bx+ay$.

Hint 2:

Integrate first over $v$ and then over $u$.

Hint 2 howto:

We continue $$\cdots = \INT_{-1}^1 \dd u \INT_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\dd v\, \frac{1}{\sqrt{1-u}} = \INT_{-1}^1 \frac{2\sqrt{1-u^2}}{\sqrt{1-u}}\,\dd u.$$

Hint 3:

This is an easy integral.

Hint 3 howto:

We finish $$\cdots = \INT_{-1}^1 2\sqrt{1+u}\,\dd u = \frac{8\sqrt2}{3}.$$


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