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Evaluate $\int_{-1}^{1} \frac 1x dx$

Does this integral converge or diverge because if we "just solve it" we get

$$\int_{-1}^1 \frac 1x dx = \ln \left| x\right|_{-1}^1=0.$$

But if we do this: $\int_{-1}^0\frac 1x dx + \int_0^1 \frac 1x dx$ it diverges. Why? What really happens here?

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    $\begingroup$ does it helps? en.wikipedia.org/wiki/Cauchy_principal_value .I think the integral is left undefined but if you use Cauchy principal value it is$0$. $\endgroup$ – Tony Ma Jun 23 '18 at 11:08
  • $\begingroup$ As with all integrals, the "limits" doesn't have to converge at the same speed. $\int_{-1}^0\frac{1}{x}\text{d}x$ and $\int _{0}^1\frac{1}{x}\text{d}x$ diverge, so the integral is divergent. But there is "Cauchy's principal value" as stated earlier, when the limits converge at the same speed. $\endgroup$ – Jakobian Jun 23 '18 at 11:13
  • $\begingroup$ take a textbook that have theory of improper integrals. Indeed there is a book titled Improper Riemann Integrals that cover many topics about these integrals. In particular it shows that, in general, it is a mistake to use a "primitive" as you did in you false identity. $\endgroup$ – Masacroso Jun 23 '18 at 11:15
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    $\begingroup$ It's worth pointing out, since no-one has yet, that the connection between integrals (area under curves) and differentiation is a consequence of the Fundamental Theorem of Calculus. The particular theorem being used here requires that the integrand, $\frac{1}{x}$, be continuous over the domain, in this case, $[-1, 1]$. The great big asymptote in the middle of this interval suggests that the FTC simply doesn't apply, and any conclusions you get are erroneous. $\endgroup$ – Theo Bendit Jun 23 '18 at 11:23
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    $\begingroup$ Technically, the indefinite integral / antiderivative / primitive of $\frac1x$ isn't $\ln |x|$. It isn't even $\ln |x|+C$. It's $\ln|x|+C(x)$ where $C(x)=C_1$ for $x<0$ and $C(x)=C_2$ for $x>0$, for two real numbers $C_1,C_2$. $\endgroup$ – Arthur Jun 23 '18 at 11:24
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You cannot apply the fundamental theorem of calculus since the function $$\frac{1}{x}$$ is not defined on the interval $[-1,1]$. Split the integral in the following way $$\int_0^1\frac{dx}{x}+\int_{-1}^0\frac{dx}{x}$$

Now you can apply the following theorem

Let $a$ be a real, then $$\int_0^a \frac{1}{x^\alpha}dx$$ converges if and only if $\alpha>1$.

Hence both integral diverge, then the sum of them diverges.

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This is an improper integral. It is defined as$$\lim_{\varepsilon\to0^+}\int_\varepsilon^1\frac1x\,\mathrm dx+\lim_{\varepsilon\to0^-}\int_{-1}^\varepsilon\frac1x\,\mathrm dx$$if both limits exist. In this case, none of the limits exist (in $\mathbb R$).

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For $[a,b]$ an interval in $\mathbb{R}$, let $f\colon [a,b]\to \mathbb{R}$. Then $F\colon [a,b]\to \mathbb{R}$ is an antiderivative of $f$ on $[a,b]$ if $F$ is continuous on $[a,b]$. It will be necessarily so in $(a,b)$, while left continuous at $a$: $F(a)=\lim_{x\to a^+} F(x)$, and right continuous at $b$: $F(b)=\lim_{x\to b^-} F(x)$ .

If $f\colon [a,b]\to \mathbb{R}$ has an antiderivative $F$ on $[a,b]$, then $f$ is integrable and so we have the Fundamental Theorem of Calculus: $$\int_a^b f(x)\,dx=F(b)-F(a)$$

The integral in the question is improper in the sense that the function $\frac{1}{x}$ is unbounded over the limits of integration $[-1,1]$, namely at the problem spot $x=0$ where it cannot be continuously defined, with the branch of the hyperbola to the left of $x=0$ an asymptote to negative $y$-axis, the branch of the hyperbola to the right of $x=0$ an asymptote to positive $y$-axis. Therefore you cannot invoke the Fundamental Theorem of Calculus over $[-1,1]$ in the manner you have done as you integrate over a discontinuity.

As an example of something that looks like it shouldn't integrate but does consider $g(x)=\frac{\sin x}{x}$; at the singular point $x=0$ we have the indeterminate expression $\frac{0}{0}$, but we know $$\lim_{x\to0} \frac{\sin x}{x}=1$$ and so $g(x)$ is continuous, having $g(0)=1$, and is bounded over $[-1,1]$.

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