$\ln(y-3) = \ln(3-y)$? How?...

Question

I am working on the same question as this one someone asked on yahoo answers...

https://uk.answers.yahoo.com/question/index?qid=20101207140137AAK5c1v

$$\int^2_1\frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \, dy.$$

Now the majority of the question I managed myself. The one issue was the $\dfrac{1}{5}\ln(y-3)$ produced from the third partial fraction $\dfrac{C}{y-3}$ Since the interval is $[1,2]$ , $\ln(y-3)$ will be undefined for both when coming to work out the definite intergral.

The best answer said to "switch $(y-3)$ to $(3-y)$". Had a similar idea, though I thought, surely $(y-3) = -(3-y)$. Meaning, I'd have $-\dfrac{1}{5}\ln(3-y)$ instead of $\dfrac{1}{5}\ln(y-3)$ So I used that but ended up with some log functions I couldn't simplify. The answerer also got the same answer that's in the back of the book.

This suggests that I did it incorrectly, but I don't see how? Anyone care to clear this up for me?

Answer 1

The correct entry in your list of standard integrals should be $$ \int \frac{\mathrm dx}x=\ln|x|+C$$ (and even this form has problems if you use both negative and positive $x$).

Answer 2

Since my old answer has gotten 2 downvotes, and the comments explain pretty much what people found objectionable about my presentation, I have rewritten it, hopefully to present my ideas in a more acceptable fashion. I have separated my answer into two sections; one for a real analysis approach, and another for a complex analysis approach.

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Real Analysis

When $y\gt3$, we have $\frac{\mathrm{d}}{\mathrm{d}y}\log(y-3)=\frac1{y-3}$. Therefore, for $y\gt3$, we have $$ \int\frac{\mathrm{d}y}{y-3}=\log(y-3)+C\tag{1} $$ When $y\lt3$, we have $\frac{\mathrm{d}}{\mathrm{d}y}\log(3-y)=\frac1{y-3}$. Therefore, for $y\lt3$, we have $$ \int\frac{\mathrm{d}y}{y-3}=\log(3-y)+C\tag{2} $$ For convenience, we can combine $(1)$ and $(2)$ into $$ \int\frac{\mathrm{d}y}{y-3}=\log|y-3|+C\tag{3} $$ However, $(3)$ must be handled with a bit of care: it cannot be applied to a definite integral with limits on both sides of $3$. For example, $$ \int_2^4\frac{\mathrm{d}y}{y-3}\tag{4} $$ does not converge. This is because neither $$ \int_2^3\frac{\mathrm{d}y}{y-3}\qquad\text{nor}\qquad\int_3^4\frac{\mathrm{d}y}{y-3}\tag{5} $$ converge. However, $(4)$ has a Cauchy Principal Value of $0$.

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Complex Analysis

As a function on $\mathbb{C}$, $\log|y-3|$ is not differentiable. Therefore, while $(1)$ and $(2)$ still hold, unfortunately, $(3)$, the convenience form for paths in $\mathbb{R}\!\setminus\!\{3\}$, does not.

When restricted to $\mathbb{R}$, any path from $2$ to $4$ must pass through $3$ and we run into the problems presented in $(4)$ and $(5)$. However, in $\mathbb{C}$, there are many paths from $2$ to $4$ that do not pass through $3$. For example, the path $\gamma_1:[0,1]\to\mathbb{C}$ given by $\gamma_1(t)=3-e^{\pi it}$:

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If we integrate along $\gamma_1$, we get $$ \begin{align} \int_{\gamma_1}\frac{\mathrm{d}y}{y-3} &=\int_0^1\frac{-\pi i\,e^{\pi it}\,\mathrm{d}t}{-e^{\pi it}}\\ &=\pi i\tag{6} \end{align} $$ By Cauchy's Integral Theorem, any path from $2$ to $4$ that does not pass over $3$ will give the value $\pi i$.

However, we also have the path $\gamma_2:[0,1]\to\mathbb{C}$ given by $\gamma_2(t)=3-e^{-\pi it}$:

$\hspace{5cm}$

If we integrate along $\gamma_2$, we get $$ \begin{align} \int_{\gamma_2}\frac{\mathrm{d}y}{y-3} &=\int_0^1\frac{\pi i\,e^{-\pi it}\,\mathrm{d}t}{-e^{-\pi it}}\\ &=-\pi i\tag{7} \end{align} $$ By Cauchy's Integral Theorem, any path from $2$ to $4$ that does not pass under $3$ gives the value $-\pi i$.

In fact, the integral of $\frac1{y-3}$ along any path between two points, $a$ and $b$, will differ by an integer multiple of $2\pi i$ from the integral along any other path between $a$ and $b$. This is because of the residue of $\frac1{y-3}$ at $y=3$. To have a well-defined anti-derivative of $\frac1{y-3}$, we need to make a branch cut that prevents a path from circling $3$.

For example, if we make the branch cut along $3+i[0,\infty)$, we allow $\gamma_1$. In this case, for $y\in\mathbb{R}$, $$ \log(3-y)-\log(y-3)=\left\{\begin{array}{} -\pi i&\text{if }y\gt3\\ +\pi i&\text{if }y\lt3 \end{array}\right.\tag{8} $$ and the two anti-derivatives differ by a constant (which can be incorporated into the constant of integration).

However, if we make the branch cut along $3-i[0,\infty)$, we allow $\gamma_2$. In this case, for $y\in\mathbb{R}$, $$ \log(3-y)-\log(y-3)=\left\{\begin{array}{} +\pi i&\text{if }y\gt3\\ -\pi i&\text{if }y\lt3 \end{array}\right.\tag{9} $$ and the two anti-derivatives differ by a constant (which can be incorporated into the constant of integration).

With such a branch cut, Cauchy's integral theorem guarantees $\int\frac{\mathrm{d}y}{y-3}$ can be well-defined on the rest of $\mathbb{C}$.

Answer 3

Just to answer to the title of the question, "$\ln(y-3) = \ln(3-y)$" is never correct. I wouldn't say it is false (it does not become correct by changing the equals sign to an not-equals sign), it just never makes sense because there are no values $y$ for which both sides are defined.

(Unless one is referring to the complex logarithm function, but (1) that is clearly not the case in this problem about integration over a real interval, (2) one needs to choose a branch of the logarithm function that makes both sides defined, and (3) after that the equality never holds, since $0$ is not among the many possible values for $\ln(-1)$.)