# Improper Riemann integral questions

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!

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?
• 4. Oh of course then we get $2\int_{0}^{1} \frac{1}{\sqrt{x}} \; dx=4$?. – squenshl Aug 4 at 3:30