Is it possible? improper integral no answer?

Question

Is it possible to evaluate this improper integral? If not then how should I answer?? $$ \int_{0}^1 \frac1x dx $$

Answer 1

Using the definition of improper integral, we have that

$$\int_0^1 \frac{1}{x}dx =\lim_{t\to0^+}\int_t^1\frac{1}{x}dx =\lim_{t\to0^+}\ln x|_t^1$$

We can see that the limit tends to infinity as $t$ tends to $0^+$.

Answer 2

I'd just say the integral diverges. You can argue using a limit or mention the singularity near $0$.

Problems of the form $\int_0^\infty \frac{1}{x^p} dx$ can be generalized to two cases, namely: $$\int_0^1 \frac{1}{x^p} dx \tag{1}$$ $$\int_1^\infty \frac{1}{x^p} dx \tag{2}$$

$(1)$ converges if and only if $p \lt 1$ and $(2)$ converges if and only if $p \gt 1$. Keep in mind that $$\int_0^\infty \frac{1}{x^p} dx \tag{3}$$

diverges for all $p \in \mathbb{R}$.

Answer 3

For $t >0$ we have $\int_t^1 {1 \over x} dx = -\ln t$, and $\lim_{t \downarrow 0} (-\ln t) = \infty$.

Since $\int_0^1 {1 \over x} dx \ge \int_t^1 {1 \over x} dx$ for all $t \in (0,1)$, we must have $\int_0^1 {1 \over x} dx = \infty$.