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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.

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

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  • $\begingroup$ This video actually helps a lot, thank you. $\endgroup$ – William Slater Sep 21 '19 at 18:07
  • $\begingroup$ It is Andrew who posted the link to the video. $\endgroup$ – zwim Sep 21 '19 at 18:09
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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 $1.$

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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.

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