Inspired by Pranjal Jain

The integral does not converge.

When integrating over a vertical asymptote, we have to be very careful that the integral converges absolutely. We cannot simply say "the graphs are symmetric hence the parts must cancel out".

In this case, because $\int_{-1} ^ 1 \left| \frac{ 1}{x} \right| \, dx$ is not finite, hence the integral in the question is undefined.

Note: $\int_{-1} ^ 1 \frac{ sgn(x) }{ x^{1/2} } \, dx = 0$ .

The function $\frac{1}{x}$ has a discontinuity at $x = 0$ ; it decreases without bound as we approach $0$ from the left and increases without bound as we approach $0$ from the right. Thus, we have to split up the integral at the discontinuity and use limits to evaluate:

$\int_{-1}^1 \frac{1}{x} dx = \lim_{a \to 0^-} \int_{-1}^a \frac{1}{x} dx + \lim_{b \to 0^+} \int_b^1 \frac{1}{x} dx$

The antiderivative of $\frac{1}{x}$ is $\ln |x|$ :

$= \lim_{a \to 0^-} \ln |x|\Big|_{-1}^a + \lim_{b \to 0^+} \ln |x|\Big|_b^1$ $= \lim_{a \to 0^-} (\ln (-a) - \ln 1) + \lim_{b \to 0^+} (\ln 1 - \ln b)$ $= \lim_{a \to 0^-} \ln (-a) - \lim_{b \to 0^+} \ln b$

Since these limits don't exist, the integral can't be evaluated.

This is explanation of the answer given by #Calvin Lin

Fundamental theorem of calculus says that differentiation is reverse process for integration only for the functions which are continuous in the interval. Since 1/x is discontinuous at x=0 , we can't integrate it using Fundamental Theorem of Calculus.

Now, we have to go to basics to find the answer.

If we divide the interval in infinitesimal rectangles, we have to see the area bounded by the rectangle as x approaches 0.

If we can neglect that area bounded by that rectangle, we can simply use Fundamental Theorem of calculus. (Reason: If we can neglect area bounded if x tends to 0, we can remove the point 0 from our domain, and if we remove point from our domain, we can use fundamental theorem of calculus).

A simple way to do this is to first to integrate it within the given limit without caring about any rule. If answer is finite, then only we can neglect that area. If the answer is infinite, then we can't neglect the area bounded by that rectangle. And if we can neglect the area bounded by that curve, we can use fundamental theorem of Calculus.

Reason why we can neglect the area when answer is finite:

If you know basics about convergent and divergent series, then it is easier to observe why we can do so. If we remove some terms from a convergent series, then the new generated series is also convergent and summation of a convergent series is always finite (more precisely,when number of terms will tend to infinity, then it will approach a finite limit).

Interestingly, if number of terms are infinite, then it doesn't matter how many finite terms you neglect, you will get the same answer. There are many more interesting properties of convergent and divergent series. I will post a question regarding this.

While if the series is divergent, Then it's summation will be infinite(more precisely,when number of terms will tend to infinity, then it will approach a infinite limit). Example of a divergent series is summation of terms in Harmonic Progression. Some Divergent series can be changed to convergent series by removing some finite terms. So, we can't neglect any term in a divergent series.

Here, we have removed the area bounded by the rectangle when x approaches 0. Here we have applied Fundamental Theorem of calculus, and got a infinite answer, then we can simply state that the initial series was divergent and if above series was divergent, we can't neglect the area bounded by that interval. And also,we can conclude nothing about area bounded by the rectangle as x approaches infinity. Hence, the integral doesn't exists.

The way I look at it, $\frac{1}{x}$ has a discontinuity at zero, hence integrating on $[-1,1]$ doesn't make sense, so it gets rewritten as two separate integrals: $\lim_{a \to 0^-} \int_{-1}^{a} \frac{1}{x} \ dx + \lim_{b \to 0^+} \int_{b}^{1} \frac{1}{x} \ dx = \left((\ln(0^-)-\ln(-1)) + (\ln(1)-\ln(0^+)) \right)$ $= -\infty + \infty$ Which doesn't exist.

it is obvious that the integral of 1/x w.r.t x is lnx whose asymptote is y=0, bt the interval is from -1 to 1 or twice the range from 0 to 1 and fortunately this does help us because we will have 2[ln1-ln0]=2ln1-2ln0=0-2ln0=-2ln0. the question is, what power of e will give zero? i can safely say the power should be of form -x for some pstive x so that e^-x will approach zero for some pstive x, the good news is it will not be zero, yep thus right. and so x must approach infinity and in turn, our desired results:it does not exist

Without any intuition as to the properties of the asymptotes on the curve y = 1/x, by simply integrating we reach the solution: ln(1) - ln(-1) We know that the natural log of 1 is 0 but the natural log of any negative number is undefined. Hence we cannot assign a value in this case.

Simply, integrating 1/x gives us lnx So the definite integral equals ln1 - ln(-1) Since ln-1 cannot be defined, the answer is does not exist