Graphically, why is $\int_{0}^{1} \frac{1}{x} dx$ divergent but $\int_{0}^{1} \frac{1}{x^{0.999}} dx$ convergent? When the power of $x$ is less than 1, it seems that the improper integral converges. I understand the math, but I don't understand how the graphs of the two cases $\frac{1}{x}$ and $\frac{1}{x^{0.999}}$ are fundamentally different.
 A: You are right the graphs aren't fundamental different, but the integral of $\frac{1}{x^{0.99999}}$ from $0$ to $1$ is a large number and if you take an exponent, that is closer to $1$ the number gets even larger.
For demonstration let $\alpha \in (0,1)$. Then
$$
\int_{0}^1 \frac{1}{x^\alpha}\,\mathrm{d}x
= \int_{0}^1 x^{-\alpha}\,\mathrm{d}x
= \frac{1}{1-\alpha} {x^{1-\alpha}}\,\bigg\vert_0^1 = \frac{1}{1-\alpha}
$$
This shows that the integral will grow arbitrarily large, if $\alpha$ gets very close to $1$. So there is nothing contra intuitive about that.
A: We can integrate to see why.
$$\int_{0}^{1} \frac{1}{x} = \lim_{a\to0} \ln(x)\Big|_a^{1} = \lim_{a\to0} -\ln(a)$$
This value clearly diverges. However, for $x^{0.999}$, we find that
$$\int_{0}^{1} \frac{1}{x^{0.999}} = 1000x^{0.0001}\Big|_0^{1} = 1000$$.
The reason for divergence is the special integral of $\frac{1}{x}$ compared to other 'powers'.
Edit: The graphs are not fundamentally different; the reason for divergences is more analytic rather thang graphical/geometric.
