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")
![\[ \sum_{n=1}^{\infty} \left( 1 - n \sin \frac{1}{n} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a6ae6110a423c01ff8e4f7fbc35bf564_l3.png "Rendered by QuickLaTeX.com")
![\[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b194923a265468a6093be62026b6f933_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{1 - n \sin \frac{1}{n}}{\frac{1}{n} \sin \frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} \frac{\frac{1}{n} - \sin \frac{1}{n}}{\frac{1}{n^2} \sin \frac{1}{n}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7ba1e2bcc4e5f7a820f79c26dd43efe2_l3.png "Rendered by QuickLaTeX.com")
, and use L’Hopital’s rule three times to take the limit,![\begin{align*} \lim_{n \to \infty} \frac{\frac{1}{n} - \sin \frac{1}{n}}{\frac{1}{n^2} \sin \frac{1}{n}} &= \lim_{x \to 0} \frac{x - \sin x}{x^2 \sin x} \\[9pt] &= \lim_{x \to 0} \frac{1 - \cos x}{2x \sin x + x^2 \cos x} \\[9pt] &= \lim_{x \to 0} \frac{sin x}{(2+x^2) \sin x + 4x \cos x} \\[9pt] &= \lim_{x \to 0} \frac{\cos x}{-2x \sin x + (6+x^2)\cos x} \\[9pt] &= \frac{1}{6}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f0f56437b41e97cea944ed0be6b41b9e_l3.png "Rendered by QuickLaTeX.com")
converges absolutely we have established the absolute convergence of![\[ \sum_{n=1}^{\infty} \left( 1 - n \sin \frac{1}{n} \right). \qquad \blacksquare\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-38e182145ca5fc9331f53b1b216a4041_l3.png "Rendered by QuickLaTeX.com")
so that the following formula is true,![\[ \lim_{x \to +\infty} \left( \frac{x+c}{x-c} \right)^x = 4. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8876362fccac848a8d5e218d796be9c5_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} &&\lim_{x \to +\infty} \left( \frac{x+c}{x-c} \right)^x &= 4 \\[9pt] \implies && \lim_{x \to +\infty} e^{ x (\log (x+c) - \log(x-c))} &= 4\\[9pt] \implies && \exp \left( \lim_{x \to +\infty} x (\log(x+c) - \log(x-c)) \right) &= 4 \\[9pt] \implies && \lim_{x \to +\infty} x(\log(x+c) - \log(x-c)) &= \log 4 \\[9pt] \implies && \lim_{x \to +\infty} \frac{\log(x+c) - \log(x-c)}{\frac{1}{x}} &= 2 \log 2 \\[9pt] \implies && \lim_{x \to +\infty} \frac{ \frac{1}{x+c} - \frac{1}{x-c}}{-\frac{1}{x^2}} &= 2 \log 2 \\[9pt] \implies && \lim_{x \to +\infty} \left( \frac{x^2}{x-c} - \frac{x^2}{x+c} \right) &= 2 \log 2 \\[9pt] \implies && \lim_{x \to +\infty} \frac{x^3 + cx^2 - x^3 + cx^2}{(x-c)(x+c)} &= 2 \log 2 \\[9pt] \implies && \lim_{x \to +\infty} \frac{ 2cx^2}{x^2 - c^2} &= 2 \log 2 \\[9pt] \implies && \lim_{x \to +\infty} \frac{2c}{1 - \left( \frac{c}{x} \right)^2} &= 2 \log 2 \\[9pt] \implies && 2c &= 2\log 2 \\[9pt] \implies && c &= \log 2. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a6e96ceff1ba5235a2bd3bcf8a811501_l3.png "Rendered by QuickLaTeX.com")
![\[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-47170b9bc0b1512b725e47ea6c62a6a0_l3.png "Rendered by QuickLaTeX.com")
and then apply L’Hopital’s rule twice. (Applying L’Hopital’s is going to be a challenge since the derivatives are going to get quite messy before we get anywhere.) We start by putting things over a common denominator,![\[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log (1+x)} \right) &= \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c2e48cfa51be106806c41ffbeb2b019e_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} D \left( \log (1+x) + \log \left( x + \sqrt{1+x^2} \right) \right) &= \frac{1}{1+x} + \frac{1}{x+\sqrt{1+x^2}} \cdot \left( 1 + \frac{x}{\sqrt{1+x^2}} \right) \\[9pt] &= \frac{1}{1+x} + \frac{ 1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt] &= \frac{1}{1+x} + \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2} \left( x + \sqrt{1+x^2} \right)} \\[9pt] &= \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-efef6a3a752cf7757d35335730956e68_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} D \left( \log \left( 1 + \sqrt{1+x^2} \right) \log (1+x) \right) &= \frac{1 + \frac{x}{\sqrt{1+x^2}}}{1+\sqrt{1+x^2}} \cdot \log(1+x) + \frac{1}{1+x} \cdot \log \left( 1 + \sqrt{1+x^2} \right) \\[9pt] &= \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-61ba70df0bd4c4631c2fa17213e07a4a_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right) &= \lim_{x \to 0} \frac{ \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}}{ \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}} } \\[9pt] &= \lim_{x \to 0}\frac{ (1+x)\sqrt{1+x^2} \left( \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}} \right)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)} \\[9pt] &= \lim_{x \to 0} \frac{ \sqrt{1+x^2} - (1+x)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-162f458c14f49b528e2589450c36546a_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} D \left( \sqrt{1+x^2} - (1+x) \right) &= \frac{x}{\sqrt{1+x^2}} -1 \\[9pt] &= \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5c263d935d646b019e8f6efcce3008a0_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} D &\left( (1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right) \right) \\[9pt] &= \log(1+x) + (1+x) \frac{1}{1+x} + \frac{x}{\sqrt{1+x^2}} \log \left( x + \sqrt{1+x^2} \right) + \sqrt{1+x^2} \frac{1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt] &= 2 + \log(1+x) + \frac{x \log \left( x + \sqrt{1+x^2} \right)}{\sqrt{1+x^2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3cedcbb3064cd67bc4f63c07ae191cb2_l3.png "Rendered by QuickLaTeX.com")
so the limit is the value of the function at ![\[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right) = -\frac{1}{2}.\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8091cf36c06c29d4f994150805d9a990_l3.png "Rendered by QuickLaTeX.com")
![\[ \lim_{x \to 1} (2-x)^{\tan \left( \frac{\pi x}{2} \right) }. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f89dc1a97a2ab0b1bbe23238d017063a_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 1} (2-x)^{\tan \left( \frac{\pi x}{2} \right)} &= \lim_{x \to 1} e^{ \tan \left( \frac{\pi x}{2} \right) \cdot \log (2-x) } \\[9pt] &= \exp \left( \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) \log (2-x) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-93bcd2371006d94373e75518ad03b87c_l3.png "Rendered by QuickLaTeX.com")
to rewrite the term in the limit in the indeterminate form ![\begin{align*} \exp \left( \lim_{x \to 1} \tan \left( \frac{\pi x}{2} \right) \log (2-x) \right) &= \exp \left( \lim_{x \to 1} \frac{\log (2-x)}{\cot \frac{\pi x}{2} } \right) \\[9pt] &= \exp \left( \lim_{x \to 1} \frac{-\frac{1}{2-x}}{-\frac{\pi}{2} \frac{1}{\sin^2 \frac{\pi x}{2}}} \right) \\[9pt] &= \exp \left( \lim_{x \to 1} \frac{2 \sin^2 \left( \frac{\pi x}{2} \right)}{\pi(2-x)} \right) \\[9pt] &= e^{\frac{2}{\pi}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6a4295a4f35c59789963b145131b84e9_l3.png "Rendered by QuickLaTeX.com")
![\[ \lim_{x \to 0^+} x^{\frac{e}{1 + \log x}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-138181e23113d1076a019cec57c435a5_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 0^+} x^{\frac{e}{1 + \log x}} &= \lim_{x \to 0^+} e^{ \frac{e}{1+\log x} \log x } \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{e \log x}{1 + \log x} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{\frac{e}{x}}{\frac{1}{x}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} e \right) \\[9pt] &= e^e. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cba7d321699673f48a3e57c50106bf2e_l3.png "Rendered by QuickLaTeX.com")
![\[ \lim_{x \to 0^+} \left( \log \left( \frac{1}{x} \right) \right)^x. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2c018e35843fc3ced0790ad1b376efe1_l3.png "Rendered by QuickLaTeX.com")
![\[ \log \left( \frac{1}{x}\right) = \log x^{-1} = - \log x. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6c0de26527f1dcd6993956b16e44c953_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 0^+} \left( \log \left( \frac{1}{x} \right) \right)^x &= \lim_{x \to 0^+} (-\log x)^x \\[9pt] &= \lim_{x \to 0^+} e^{- x \log (\log x)} \\[9pt] &= \exp \left( \lim_{x \to 0^+} (-x \log (\log x)) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \log (\log x)}{\frac{1}{x}} \right)\\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ -\frac{1}{x \log x}}{-\frac{1}{x^2}} \right) & (\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{x}{\log x} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{1}{\frac{1}{x}} \right)&(\text{L'Hopital}) \\[9pt] &= e^0 = 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-228fa614f71767336ed91f839b92b116_l3.png "Rendered by QuickLaTeX.com")
![\[ \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0045ebd98cfcea4c46589af011999c1c_l3.png "Rendered by QuickLaTeX.com")
to write,![\begin{align*} \lim_{x \to 0^-} \left( 1 - 2^x \right)^{\sin x} &= \lim_{x \to 0^-} e^{\sin x \log (1-2^x)} \\[9pt] &= \exp \left( \lim_{x \to 0^-} \left( \sin x \log (1-2^x) \right) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0a94819bcf7da4633fa0fdc06e681622_l3.png "Rendered by QuickLaTeX.com")
and apply L’Hopital’s rule,![\begin{align*} \exp &\left( \lim_{x \to 0^-} \left( \sin x \log (1-2^x) \right) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{\log (1-2^x)}{\frac{1}{\sin x}} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{-\frac{2^x \log 2}{1-2^x}}{-\frac{\cos x}{\sin^2 x}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^-} \frac{ \sin^2 x \cdot 2^x \log 2}{\cos x \cdot (1-2^x)} \right) \\[9pt] &= \exp \left( \log 2 \cdot \lim_{x \to 0^-} \left( \frac{2^x \sin x}{(1-2^x) \cos x} \right) \right) \\[9pt] &= \exp \left( \log 2 \cdot \lim_{x \to 0^-} \left( \frac{ 2^{x+1} \sin x \cos x + 2^x \log 2 \sin^2 x}{-2^x \cos x \log 2 - (1-2^x) \sin^2 x} \right) \right) &(\text{L'Hopital}). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0620401ac551d0ea8640a5a83dfb7127_l3.png "Rendered by QuickLaTeX.com")
is equal to the value of the function. Since the numerator is 0 and the denominator is 
when 
![\[ \lim_{x \to 0^+} x^{\left( x^x - 1 \right)}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-319e927fac61cd591e37f440ca960a09_l3.png "Rendered by QuickLaTeX.com")
), and use the continuity of the exponential function. Then we apply L’Hopital’s rule twice. (We also use the notation 
to avoid huge expressions in the exponent.)![\begin{align*} \lim_{x \to 0^+} x^{\left( x^x - 1 \right)} &= \lim_{x \to 0^+} e^{\left(x^x -1 \right) \log x} \\[9pt] &= \exp \left( \lim_{x \to 0^+} \left((x^x-1) \log x \right)\right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \log x}{\frac{1}{x^x-1}} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \frac{1}{x}}{ \frac{x^x (\log x + 1)}{(x^x-1)^2}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{(x^x-1)^2}{x^x(x \log x + x)} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{2(x^x-1) x^x (\log x + 1)}{x^x (x \log x + x)(\log x + 1) + x^x(\log x+ 2)}\right) &\text{L'Hopital} \\[9pt] &= \exp \left( \lim_{ x \to 0^+} \frac{2(x^x-1)(\log x + 1)}{(x \log x + x)(\log x + 1) + \log x + 2} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{2(x^x -1)}{x(\log x + 1) + 1 + \frac{1}{\log x + 1}} \right) \\[9pt] &= e^0 \\ &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4e851db3e01d29c9f82f1104b7cbb1a5_l3.png "Rendered by QuickLaTeX.com")
by Example #3 on page 302 of Apostol, which means the numerator is going to 0 as 
. However, in the denominator both the terms 
and 
go to 0 as 
in the denominator means the denominator is going to 1 as ![\[ \lim_{x \to 1^-} \big( (\log x)(\log (1-x)) \big). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f26426c70d1043667acb1c607cb026de_l3.png "Rendered by QuickLaTeX.com")
and apply L’Hopital’s rule,![\begin{align*} \lim_{x \to 1^-} \big( (\log x)(\log (1-x)) \big) &= \lim_{x \to 1^-} \frac{\log (1-x)}{\frac{1}{\log x}} \\[9pt] &= \lim_{x \to 1^-} \frac{\frac{-1}{1-x}}{\frac{-1}{x (\log x)^2}} \\[9pt] &= \lim_{x \to 1^-} \frac{-x (\log x)^2}{1-x} \\[9pt] &= \lim_{x \to 1^-} - \frac{(\log x)^2 + 2 \log x}{-1} &(\text{L'Hopital again}) \\[9pt] &= \lim_{x \to 1^-} \big( (\log x)^2 + 2 \log x \big) \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-98d3554788a016ae66955795e678b587_l3.png "Rendered by QuickLaTeX.com")