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

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.

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    $\begingroup$ For $x=0$ the integrand $x^{-2}$ is not defined. That means that the integral $\int_{\left[0,1\right]}x^{-2}dx$ is not defined. The integral $\int_{(0,1]}x^{-2}dx$ can be interpreted as $\lim_{\varepsilon\downarrow0}\int_{\varepsilon}^{1}x^{-2}dx=\lim_{\varepsilon\downarrow0}\left[\varepsilon^{-1}-1\right]=\infty$. $\endgroup$ – drhab Jun 12 '14 at 9:35
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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: $$\left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 \le x \le 1 \right] = \left[ \int_0^1 f(x) \, \text{d}x, \, \text{for } 0 < x \le 1 \right] + \int_0^0 f(x) \, \text{d}x$$

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.

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  • $\begingroup$ Why the downvote? $\endgroup$ – JoeyBF Jun 12 '14 at 10:30

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