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The function $f:(-\infty,0)\cup (0,\infty)\rightarrow \mathbb{R}$ is defined by $f(x) := \frac{1}{\sqrt{|x|}}$ $\left(\text{so on $(0,\infty)$ we have $f(x) = \frac{1}{\sqrt{x}}$}\right)$.
1. Determine whether $f$ is improperly Riemann integrable over $[0,1]$, and calculate $\int_{0}^{1} \frac{1}{\sqrt{x}} \; dx$ if it is.
The improper Riemann integral is the limit \begin{equation*} \begin{split} \int_{0}^{1} \frac{dx}{\sqrt{x}} &:= \lim_{\epsilon\to 0^+} \int_{\epsilon}^{1} \frac{dx}{x^{\frac{1}{2}}} \\ &= \lim_{\epsilon\to 0^+} \left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{\epsilon}^{1} \\ &= \lim_{\epsilon\to0^+} 2-\left(2\sqrt{\epsilon}\right) \\ &= 2 \end{split} \end{equation*} 2. Determine whether $f$ is improperly Riemann integrable over $[1,\infty)$, and calculate $\int_{1}^{\infty} \frac{1}{\sqrt{x}} \; dx$ if it is.
The improper Riemann integral is the limit \begin{equation*} \begin{split} \int_{1}^{\infty} \frac{dx}{\sqrt{x}} &:= \lim_{\epsilon\to\infty} \int_{1}^{\epsilon} \frac{dx}{x^{\frac{1}{2}}} \\ &= \lim_{\epsilon\to\infty} \left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{1}^{\epsilon} \\ &= \lim_{\epsilon\to\infty} \left(2\sqrt{\epsilon}\right)-2 \\ &= \infty. \end{split} \end{equation*} Hence, $f$ is not improperly Riemann integrable over $[1,\infty)$.
3. Explain why $f$ is not improperly Riemann integrable over $[0,\infty)$.
4. Is $f$ improperly Riemann integrable over $[-1,1]$? Briefly explain your answer.
Well we get $2-2i$.
Any help for the third and fourth questions will be helpful!

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  1. By additivity of the integral: $$\int_0^\infty f(x)\;dx = \int_0^1 f(x)\;dx + \int_1^\infty f(x)\;dx$$
  2. Note that $f$ is symmetric: $f(-x)=f(x)$. How can you then rewrite the integral?
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    $\begingroup$ 4. Oh of course then we get $2\int_{0}^{1} \frac{1}{\sqrt{x}} \; dx=4$?. $\endgroup$
    – squenshl
    Aug 4, 2019 at 3:30
  • $\begingroup$ @squenshl Yes, that‘s right. $\endgroup$ Aug 4, 2019 at 3:34

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