# Finding a definite integral of $(\frac{x}{2x-1})$

Once again I'm doing some preparatory work for a course and have hit a stumbling block and would appreciate some pointers. I think I'm most of the way there...

So the question is calculate the following: $$\int_0^1 \frac{x}{2x-1}$$ Which to me is definite integrals. My first thought was to express as partial fractions in order to make it easier to apply rules, but as the numerator must be a lower degree than the denominator, I must do a long division, which I did as follows:

$$x/(2x-1)$$ Which gives , simplified: $$\frac{1}{2}. \frac{1/2}{2x-1}$$ working out as: $$(\frac{1}{2})(\frac{1}{2(2x-1)})$$ Now I can apply the fact that $$\frac{1}{ax+b} = \frac{1}{a}\ln(ax+b) + C$$, AND integrate individual terms, ADN factor out common items meaning: $$(\frac{1}{2})(\frac{1}{2(2x-1)}) = \frac{1}{2}\int\frac{1}{2x-1}dx + \frac{1}{2}\int1dx$$ Which works out on paper to: $$\frac{1}{2}x + \frac{1}{4}\ln(2x-1) + C$$

But then if you attempt to substitute in your upper and lower limits, you are trying to take the natural log of zero, which is a maths error as far as I can see?

I saw a similar question here: Integral of $x/(2x-1)$ But thats not dealing with limits. Is this a trick question or something? Regardless any help would be greatly appreciated.

• What happens when $x=\frac12$? – Peter Foreman Sep 21 '19 at 17:11
• @William Slater: To better understand, take a look at this graph.desmos.com/calculator/etq6oujlc8 – Khosrotash Sep 21 '19 at 17:22
• I'm afraid I'm not sure I understand? I mean that line would be a vertical line? – William Slater Sep 21 '19 at 17:24

The problem is the $$\frac 12$$ is a pole in this integral.

First note that it should be $$\frac 12x+\frac 14\ln|2x-1|+C$$ with a different constant on any interval not containing the pole.

So you have to study $$\displaystyle I(a,b)=\int_0^{\frac 12-a}\frac x{2x-1}\mathop{dx}+\int_{\frac 12+b}^1\frac x{2x-1}\mathop{dx}$$

If $$\lim_{(a,b)\to(0^+,0^+)}I(a,b)$$ exists then your integral is well defined.

This is not the case here since you'll get some $$\ln(a)-\ln(b)$$ which diverges.

You could also examine if $$\lim_{a\to 0^+}I(a,a)$$ exists, this "symmetrical" calculation is called the Cauchy principal value.

https://fr.wikipedia.org/wiki/Valeur_principale_de_Cauchy

This time, the logarithms cancel each other and we get $$\frac 12$$ as CPV.

• This video actually helps a lot, thank you. – William Slater Sep 21 '19 at 18:07
• It is Andrew who posted the link to the video. – zwim Sep 21 '19 at 18:09

The integrand is $$\frac12\left(1+\frac{1/2}{x-1/2}\right).$$ Thus the integral is $$\frac12\left(x+\frac12\log|x-1/2|\right)+K.$$

This exists both at $$0$$ and $$1.$$

Let $$f(x)=\dfrac{x}{2x-1}.$$ The function is undefined at $$x=\frac12$$, so we must rewrite our integral as $$\int_0^1\frac{x}{2x-1}\ dx=\lim_{t\to\frac12^-}\int_0^t\frac{x}{2x-1}\ dx+\lim_{s\to\frac12^+}\int_s^1\frac{x}{2x-1}\ dx.$$

If this is for anything before Calc 2, you can simply say that this integral diverges. But the Cauchy principal value for this integral will be $$\frac12$$. Here is a video that might explain it more clearly.