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I found myself intrigued by the result of the definite integrals of $\frac{1-x}{x}$ and $\sqrt{\frac{1-x}{x}}$ respectively. Both functions are seemingly similar, with vertical asymptotes at $x=0$. However the area bounded between the two and the $x$-axis for $x=[0,1]$ is clearly different. Let $f(x)=\frac{1-x}{x}$:

                     graph of f(x) and sqrt(f(x))

and so we have: $$ \begin{align} &\int\limits^{1}_{0}f(x)\,dx=\infty\\ &\int\limits^{1}_{0}f(x)^{1/2}\,dx=\frac{\pi}{2} \end{align} $$

Now, since $\displaystyle\lim\limits_{x\to0}\,f(x)=\lim\limits_{x\to0}\,f(x)^{1/2}=\infty$ shouldn't the two both evaluate to infinity? Even though it is clear that $\int\limits^{1}_{0}f(x)^{1/2}\,dx$ will be less than $\int\limits^{1}_{0}f(x)\,dx$, how can that be that one evaluates to infinity while the other to merely $\approx1.57\!\!\ldots$?

Has this something to do with this? Or perhaps with the fact that $\ln{x}$, which forms a part of the indefinite integral for $f(x)$, isn't defined for $0$?

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    Perhaps you'd look at a simpler example of the same phenomenon, the integrals of $1/x$ and $1/\sqrt x$. Without the clutter, you might see what's going on.2012-05-16
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    @GerryMyerson Indeed, the two cases are very similar, which is also why considering the other case didn't lend me any more insight. It still doesn't make a lot of sense to me.2012-05-16
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    Intuitively, you can think of the graph of $\sqrt{\frac{1-x}{x}}$ as "hugging" the y-axis "tighter" than $\frac{1-x}{x}$ does.2013-05-23

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