# Computing $\int_{-1}^1 \frac{1}{x^2} dx$ (improper integral)

\begin{aligned} & \phantom{ {}={} } \int_{-1}^1 \frac{1}{x^2} \, dx \\ &= \int_{-1}^0 \frac{1}{x^2} dx + \int_0^1 \frac{1}{x^2} dx \\ &=\lim_{a\to 0^-} \left[ \int_{-1}^a \frac{1}{x^2} dx \right] + \lim_{b \to 0^+} \left[ \int_b^1 \frac{1}{x^2} dx \right] \\ &= \lim_{a \to 0^-} \left[ - \frac{1}{x} \bigg\rvert^{x =a}_{x=-1} \right] + \lim_{b \to 0^+} \left[ - \frac{1}{x} \bigg\rvert^{x = 1}_{x=b} \right] \\ &= \lim_{a \to 0^-} \left[ - \frac{1}{a} + 1 \right] + \lim_{b \to 0^+} \left[ - 1 + \frac{1}{b} \right] \end{aligned}

The limit doesn't exist over either piece of the domain. Hence the integral is divergent.

Is this correct?

• As far as I can see, it looks correct. – Toby Mak Aug 12 at 4:02
• Would you apply the same justification for $\int_{-1}^1\frac 1x dx$? Since it's an odd function, a symmetric integration range around $0$ yields $0$ – Andrei Aug 12 at 4:02
• @Andrei: that integral is also divergent, but it does have a Cauchy Principal Value of $0$. – robjohn Aug 12 at 4:08