# Evaluating a Discontinuous Integral

How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$

My work:

The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:

$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to

$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3}$$.

Then recall that in evaluating a discontinuous integrand, the integral is defined by the relations $$\int_a ^b f(x) = lim_{x->b^-} \int_a ^x f(x) dx$$

if $x = b$ is the discontinuous point or $$\int_a ^b f(x) = lim_{x->a^+} \int_x ^b f(x) dx$$

if $x = a$ is the discontinuous point.

With that in mind:

$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3}$$ $$\int _0 ^3\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} + \int _3 ^4\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)}$$ $$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} + \int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$

Remembering that $\int \frac{du}{u^2 +a^2} = \frac{1}{a} \arctan \left( \frac{u}{a}\right) + C,$

we get:

$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} +\int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$

equals

$$lim_{x->3^+} \int_0 ^x \frac{dx}{(x-1)^2 -(2)^2} + lim_{x->3^-} \int_x ^4 \frac{dx}{(x-1)^2 -(2)^2}$$

equals

$$lim_{x->3^+} \left( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_3 ^x + lim_{x->3^-} \left(\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_x ^4$$

equals

$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{3-2}{2}\right) \right ) + \left ( \frac{1}{2} \arctan \left( \frac{4-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$

equals

$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)- 0.2318 \right ) + \left ( \frac{\pi}{8}-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$

which makes $$\int _0 ^4\frac{dx}{x^2 - 2x - 3} = 0.1609$$

But in my book, it said there is no value of $\int _0 ^4\frac{dx}{x^2 - 2x - 3}.$

How do you prove that the integral $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ has no value?

• The integral for arctangent has a +. You have a -.
– user296602
Commented Aug 7, 2017 at 6:34
• @user296602 oh my.....I've been double crossed by my own eyesight..... Commented Aug 7, 2017 at 6:37

The indefinite integral is $$\int \frac{1}{x^2-2 x-3} \, dx=\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]+C$$ Indeed the fraction can be split in the sum of two fractions

$\dfrac{1}{(x-3) (x+1)}=\dfrac{a}{x-3}+\dfrac{b}{x+1}=\dfrac{(a+b) x+a-3 b}{(x-3) (x+1)}$

which is equal to the given fraction if

$\left\{ \begin{gathered} a + b = 0 \hfill \\ a - 3b = 1 \hfill \\ \end{gathered} \right.\rightarrow$ $\left\{ \begin{gathered} a = \frac{1}{4} \hfill \\ b = - \frac{1}{4} \hfill \\ \end{gathered} \right.\rightarrow \dfrac{1}{(x-3) (x+1)}=\dfrac{1}{4}\left(\dfrac{1}{x-3}-\dfrac{1}{x+1}\right)$

If we write the given definite integral as you did $$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3}=\\=\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]_0^3+\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]_3^4$$ we see that the first tends to $+\infty$ as $x\to 3$ because of the logarithm

So the integral diverges

Hope this helps

• I think $\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3}=\\=\frac{1}{4} \left[\ln (x-3)-\ln(x+1)\right]_0^3+\frac{1}{4} \left[\ln (x-3)-\ln (x+1)\right]_3^4$....... Commented Aug 8, 2017 at 2:51

In fact, you should use the expression for integral of the function $\dfrac{1}{x^2-a^2}$: $$\int_3^4 \frac{dx}{(x-1)^2-2^2} = \int_2^3 \frac{dx}{x^2-2^2} = \frac{1}{4}\lim_{x\to 2^-}\left (\log\frac{1}{5} -\log\frac{x-2}{x+2}\right ).$$ The above limit does not exist.

$$I=\int _0 ^4\frac{dx}{x^2 - 2x - 3} =\frac{1}{4} \int _0 ^4\left(\frac{1}{x - 3}- \frac{1}{x+1}\right)dx$$ As already pointed out, there is a singularity at $x=3$.

So, at the common sens, the integral isn't convergent. But this is not true in the sens of Cauchy integration (Cauchy Principal Value : http://mathworld.wolfram.com/CauchyPrincipalValue.html )

$$I=\frac{1}{4}\lim_{\epsilon\to 0} \left( \int _0 ^{3-\epsilon}\frac{dx}{x - 3} +\int _{3+\epsilon}^4\frac{dx}{x - 3} - \int _0^4\frac{dx}{x+1}\right)$$

$\int _0 ^{3-\epsilon}\frac{dx}{x - 3} +\int _{3+\epsilon}^4\frac{dx}{x - 3} = \left[\ln|x-3| \right]_{x=0}^{x=3-\epsilon} +\left[\ln|x-3| \right]_{x=3+\epsilon}^{x=4} = \ln(\epsilon)-\ln(3)+\ln(1)-\ln(\epsilon) = -\ln(3)$

$\int _0^4\frac{dx}{x+1}=\left[\ln|x+1| \right]_{x=0}^4=\ln(4+1)-\ln(1)=\ln(5)$ $$\text{PV}\int _0 ^4\frac{dx}{x^2 - 2x - 3} = \frac{1}{4}\left(-\ln(3)-\ln(5)\right)=-\frac{\ln(15)}{4}$$ The symbol PV means that the integral is considered on the sens of a Cauchy integral. Indeed, the integral as a finite value, the so called Principal Value.

• So you mean, in our eyes, the integral is divergent. But in Cauchy's eyes, it is convergent? Commented Aug 8, 2017 at 2:30
• Of course. I am not the only one thinking so. All this is well-known by mathematicians since the 19th century. Commented Aug 8, 2017 at 6:22