Calculating integral with different boundary I am confused about calculting such integral with two ways of choosing boundary: 
$$I_1 = \int_0^1 \frac 1{x^2}dx,\quad \text{for } 0 \le x \le 1$$
$$I_2 = \int_0^1 \frac 1{x^2}dx, \quad \text{for } 0 &lt; x \le 1$$

What is the difference between $I_1$ and $I_2$ when we calculate these integral including the case $x=0$ and not? Thank you for your answers.
The expression $$I = \int_0^1 \frac{1}{x^2}\text{d}x$$ would contain enough information to unambiguously evaluate the integral, if it converged. This is because integrals over single points evaluate to 0.

To see this clearly, consider splitting the expression:
$$\left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 \le x \le 1 \right] = \left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 &lt; x \le 1 \right] + \int_0^0 f(x) \, \text{d}x$$

However, both diverge when $f(x) = \frac{1}{x^2}$ because the closure of the interval of integration contains an essential singularity of the integrand.
The expression $$I = \int_0^1 \frac{1}{x^2}\text{d}x$$ would contain enough information to unambiguously evaluate the integral, if it converged. This is because integrals over single points evaluate to 0.

To see this clearly, consider splitting the expression:
$$\left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 \le x \le 1 \right] = \left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 &lt; x \le 1 \right] + \int_0^0 f(x) \, \text{d}x$$

However, both diverge when $f(x) = \frac{1}{x^2}$ because the closure of the interval of integration contains an essential singularity of the integrand.