Why is $\int_{0}^{1}\frac{1}{x} dx = \int_{1}^{+\infty}\frac{1}{x} dx = +\infty$? I have the theoretical proof but I can't visualise it, doesn't it mean that the area under the function $\frac{1}{x}$ is infinite?
How is it possible since it tends towards 0?
 A: You're absolutely correct in saying that the area under the function $\frac{1}{x}$ is infinite. And yes, while the function tends to zero, it doesn't tend to zero "fast enough" for the area to converge. One possibly more intuitive way to see this is to consider the lower Riemann sum with width one. The sum of these areas is $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots$, which diverges, and since this area is less than the area of the integral, the integral must also diverge.
A: This is just a feature of infinity that you have to get used to - some sequences tend to zero too slowly for their sum (or integral) to converge. The harmonic series is a classic example of this (and in fact provides a lower bound for your integral).  
Notice that the classical paradox (cf Zeno) is that infinite series shouldn't converge at all - why would you expect that adding infinitely many (admittedly, quite small) numbers should always converge?
A: Substituting $x\mapsto at$, we get that for any $a\gt0$
$$
\int_a^{2a}\frac1x\,\mathrm{d}x=\int_1^2\frac1t\,\mathrm{d}t
$$
Furthermore, since $\frac1t\ge\frac12$ on $[1,2]$, we get
$$
\int_1^2\frac1t\,\mathrm{d}t\ge\frac12
$$
Therefore, breaking $(0,1]$ into intervals $\left(2^{-k-1},2^{-k}\right]$ for $k=0,1,2,\dots$
$$
\begin{align}
\int_0^1\frac1x\,\mathrm{d}x
&=\sum_{k=0}^\infty\int_{2^{-k-1}}^{2^{-k}}\frac1x\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\int_1^2\frac1t\,\mathrm{d}t\\
&\ge\sum_{k=0}^\infty\frac12\\[6pt]
&=\infty
\end{align}
$$
For the second integral, substitute, $x\mapsto\frac1t$:
$$
\begin{align}
\int_1^\infty\frac1x\,\mathrm{d}x
&=\int_1^0t\,\mathrm{d}\frac1t\\
&=\int_1^0t\left(-\frac1{t^2}\right)\,\mathrm{d}t\\
&=-\int_1^0\frac1t\,\mathrm{d}t\\
&=\int_0^1\frac1t\,\mathrm{d}t\\[9pt]
&=\infty
\end{align}
$$
