Evaluate the integral $ \int_{(0,1)} \frac{1}{\sqrt x}$ as an improper integral by finding a convenient exhaustion $(C_1, C_2, \cdots)$ for $(0, 1)$ and taking the limit $\int_{C_N} \frac{1}{\sqrt x}$ as $N \to \infty$.

I am unable to get such $C_N$. Help Needed.

  • 2
    $\begingroup$ The integral for $[a,b] \subset (0,1)$ is $2\sqrt{b}-2\sqrt{a}$. So just choose $a$, $b$ to be squares of some convenient sequences tending to $0$, $1$, respectively. $\endgroup$ – preferred_anon Nov 16 '17 at 13:56
  • $\begingroup$ can I take a to be 1/N and b to be 1 $\endgroup$ – user8795 Nov 16 '17 at 14:16
  • $\begingroup$ Sounds good to me! So what is the integral? (note: I would have found $1/N^{2}$ more convenient, but it is not very important) $\endgroup$ – preferred_anon Nov 16 '17 at 14:22

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