Why is it that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ is not less than $\int_1^\infty \frac{dx}{x^2} = 1$? So according to Euler's proof of the Basel problem, 
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6},$$
But only for $n  \in  \mathbb{Z}$.
But if $n$ was a positive real and $n \geqslant 1$, then would the sum $S$ be equal to,
$$\int_1^\infty \frac{dx}{x^2}?$$
If so then after solving the integral via power rule we get $S=1$.
But how can this be since when taking reals, we take integral values as well as new values. So by logic shouldn't, $S>\dfrac{\pi^2}{6}$ but $1$ isn't. Where am I going wrong pls explain.

In the following picture, the pink area is the left-hand part of the integral $\displaystyle \int_{x=1}^\infty \frac1x \, dx$ while the green and pink areas together are left-hand part of the sum $\displaystyle \sum_{n=1}^\infty \frac1{n^2}$
The green area is the difference, which is clearly positive and is in fact $\dfrac{\pi^2}{6} -1\approx 0.6449$

