Test the following improper integral for convergence:
![\[ \int_{0^+}^{1^-} \frac{dx}{\sqrt{x} \log x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0f5478e6f882c67faa88cb18760754d1_l3.png "Rendered by QuickLaTeX.com")
The integral diverges.
Proof. First, we show that 
for all 
. To do this let
![\[ f(x) = \sqrt{x} - \log x \quad \implies \quad f'(x) = \frac{1}{2 \sqrt{x}} - \frac{1}{x} = \frac{\sqrt{x} - 2}{2x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d328e585f2d1fbc746a0d8dfa1019fdb_l3.png "Rendered by QuickLaTeX.com")
This derivative is 0 at 
and is less than 0 for 
and greater than 0 for 
. Hence, 
has a minimum at . But, 
(since 
implies 
). So, has a minimum at and is positive there; thus, it is positive for all 
, or
![\[ f(x) = \sqrt{x} - \log x > 0 \quad \implies \quad \sqrt{x} > \log x \qquad \text{for all } x > 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-04c8a9a7b7808c02f4bd32751a564fc4_l3.png "Rendered by QuickLaTeX.com")
So, since we know
![\[ \frac{1}{\sqrt{x} \log x} > \frac{1}{\sqrt{x} \cdot \sqrt{x}} = \frac{1}{x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b13a62f191c39b9795fd1a9045892684_l3.png "Rendered by QuickLaTeX.com")
But then, consider the limit
![\begin{align*} \lim_{x \to 0^+} \frac{ \frac{1}{x} }{ \frac{1}{\sqrt{x} \log x}} &= \lim_{x \to 0^+} \frac{\log x}{\sqrt{x}} \\ &= \lim_{x \to 0^+} \frac{\log x}{x^{\frac{1}{2}}} \\[9pt] &= 0 &(\text{by Theorem 7.11}). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d89a3d1641438456c64130285fa86241_l3.png "Rendered by QuickLaTeX.com")
Therefore, by the limit comparison test (Theorem 10.25), the convergence of 
would imply the convergence of 
(for any 
), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of
![\[ \int_{0^-}^{1^+} \frac{1}{\sqrt{x} \log x} \, dx. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f0b0338bca443c686c99828a0fd42e31_l3.png "Rendered by QuickLaTeX.com")
Split in two, and then compare the negative lower integral with 1/x^(3/2), which converges. The upper one diverges, as can be compared with 1/x. A substitution t = 1/x could help.
RoRi, dude, log x<0 for 0<x<1 that said it isn't possible to 1/[sqrt(x) log(x)] to be greater than 1/x
You are wrong about theorem 7.11. The hypothesis of theorem 7.11 is with x is going to infinity, not to zero.
Rafael, the book says that the theorem is easily exploted to other situations, such as the x approaching 0.