As a probabilistic approximation to the Last Fermat's Theorem truthness, Feynman (physicist) wrote a simple, yet interesting, calculation based on probability. To reproduce the result it is necessary to calculate the following limit: $$\int_{0}^{\infty}\int_{0}^{\infty}(u^n+v^n)^{-1+\frac{1}{n}}\,dudv$$ when n tends to $\infty$ which is supposed to give $\frac{1}{n}$. However, my approximation was:
We write $u=rcos(\theta)$, $v=rsin(\theta)$, $dudv=rdrd\theta$, then
$$= \lim_{n \to \infty}{}\int_{0}^{\pi/2}\int_{0}^{\infty}(r^ncos^n(\theta)+r^nsin^n(\theta))^{-1+\frac{1}{n}}\,rdrd\theta$$ $$=\lim_{n \to \infty}{}\int_{0}^{\pi/2}\int_{0}^{\infty}(cos^n(\theta)+sin^n(\theta))^{-1+\frac{1}{n}}\,r^{-n}drd\theta$$ $$=\lim_{n \to \infty}{}\int_{0}^{\pi/2}(cos^n(\theta)+sin^n(\theta))^{-1+\frac{1}{n}}\,d\theta\int_{0}^{\infty}r^{-n}dr$$ $$=\lim_{n \to \infty}{}\int_{0}^{\pi/2}(cos^n(\theta)+sin^n(\theta))^{-1+\frac{1}{n}}\,d\theta\Big[\frac{r^{-n+1}}{-n+1}\Big]_{0}^\infty$$ $$=\lim_{n \to \infty}\Big[\frac{r^{-n+1}}{-n+1}\Big]_{0}^\infty\int_{0}^{\pi/2}(cos^n(\theta)+sin^n(\theta))^{-1+\frac{1}{n}}\,d\theta$$ $$=\lim_{n \to \infty}\Big[\frac{r^{-n+1}}{-n+1}\Big]_{0}^\infty\int_{0}^{\pi/2}(1+tan^n(\theta))^{-1+\frac{1}{n}}cos^{1-n}(\theta)\,d\theta$$ so when $r$ tends to $\infty$ limit goes to $0$ but when $r$ tends to $0$ limit diverges, am I missing something? Thanks in advance.