# What is the improper integral 1/(2x-1)dx from 0 to 1/2 ? . .

Jun 5, 2018

$I = {\int}_{0}^{\frac{1}{2}} \frac{1}{2 x - 1} \cdot \mathrm{dx} = {\lim}_{b \rightarrow \frac{1}{2}} {\int}_{0}^{b} \frac{1}{2 x - 1} \cdot \mathrm{dx}$

#### Explanation:

show below:

$I = {\int}_{0}^{\frac{1}{2}} \frac{1}{2 x - 1} \cdot \mathrm{dx}$

Since $\frac{1}{2 x - 1}$ not continues at $x = \frac{1}{2} \in \left[0 , \frac{1}{2}\right]$ we must use improper integral method to integrate it :

$I = {\int}_{0}^{\frac{1}{2}} \frac{1}{2 x - 1} \cdot \mathrm{dx} = {\lim}_{b \rightarrow \frac{1}{2}} {\int}_{0}^{b} \frac{1}{2 x - 1} \cdot \mathrm{dx}$

${\lim}_{b \rightarrow \frac{1}{2}} \frac{1}{2} {\int}_{0}^{b} \frac{2}{2 x - 1} \cdot \mathrm{dx}$

$\frac{1}{2} \cdot {\lim}_{b \rightarrow \frac{1}{2}} {\left[\ln \left(2 x - 1\right)\right]}_{0}^{b}$

$\frac{1}{2} \cdot {\lim}_{b \rightarrow \frac{1}{2}} \left[\ln \left(2 b - 1\right)\right]$

Note: The integral is divergent. The result shown is the Cauchy principal value

Note: It was assumed that $b > 0 , 2 b - 1 > 0$

$\frac{1}{2} \cdot {\lim}_{b \rightarrow \frac{1}{2}} \left[\ln \left(2 b - 1\right)\right] = \text{Does not exist}$