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Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist? $$ I=\frac{1}{2} \lim_{t\to 1^{-}} [\arctan(2-3/2) - \arctan(-3/2)] + \lim_{t\to 1^+}[\arctan(0/2) - \arctan(1-3/2)] $$ I can find $\arctan(0)= \pi/2$

So why the answer in my book is doesn't exist?

Michael Albanese
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user32104
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4 Answers4

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Note that $\frac{1}{(x-1)(x-5)}$ is undefined at $x = 1$ so

$$\int_0^3\frac{1}{(x - 1)(x-5)}dx = \lim_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx + \lim_{t\to 1^+}\int_t^3\frac{1}{(x-1)(x-5)}dx.$$

To evaluate the two integrals on the right, we need to find an antiderivative. We do this by first applying partial fraction. Doing so we obtain

$$\frac{1}{(x-1)(x-5)} = \frac{-\frac{1}{4}}{x-1} +\frac{\frac{1}{4}}{x-5}.$$

Therefore $-\frac{1}{4}\ln|x-1| + \frac{1}{4}\ln|x-5| = \frac{1}{4}\ln\left|\frac{x-5}{x-1}\right|$ is an antiderivative of $\frac{1}{(x-1)(x-5)}$. Therefore

$$\int_0^s\frac{1}{(x-1)(x-5)}dx = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| - \frac{1}{4}\ln\left|\frac{0-5}{0-1}\right| = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| -\frac{1}{4}\ln 5.$$

As $\lim\limits_{s\to 1^-}\frac{1}{4}\ln\left|\frac{s-5}{s-1}\right|$ does not exist, $\lim\limits_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx$ does not exist, so $\int_0^3\frac{1}{(x-1)(x-5)}dx$ does not exist.

Michael Albanese
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  • Thank you !! Is it right to say x^2-6x*5=(x-3)^2-4=4[((x-3)/2)^2-1], then x-3/2=tan@?? – user32104 Dec 30 '13 at 12:27
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    There's nothing wrong with what you've written, but I'm not sure if it will help you calculate the integral. – Michael Albanese Dec 30 '13 at 12:31
  • The fact you can establish the antiderivative and the fact that you can comute the integral are two different things. The discontinuity of the integrand in the range of integration is the origin of your problem. – Claude Leibovici Dec 31 '13 at 04:52
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The improper integral: $$\int_1^3f(x)dx$$ does not converge. In fact, using Quotient test for it gives us: $$\lim_{x\to 1^+}(x-1)^{\color{red}{1}}f(x)\neq 0$$ and so it diverges.

Mikasa
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Let $a,b,\alpha$ be real numbers such that $a<b$.

The integrals $\displaystyle \int \limits_a^{b^-} \frac{1}{(b-x)^\alpha}\mathrm dx , \int \limits_{a^+}^b \frac{1}{(x-a)^\alpha}\mathrm dx$ converge if, and only if, $\alpha <1$.

Hint: $\forall x\in \mathbb R\left((x^2-6x+5)=(x-1)(x-5)\right)$.

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Git Gud
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  • @Git thank you I already know that my problem is why after integrate this the answer in my book is infinite but I can find the answer so it's converge as you know arctan(0)=pi/2 – user32104 Dec 30 '13 at 11:58
  • As discussed in the comments in the question, the equality $\arctan(0)=\dfrac \pi 2$ is incorrect. Regarding your work, I need some help because I don't understand what you're doing. Why are you even messing with $\arctan$? Can it be that you miscalculated the antiderivatives? – Git Gud Dec 30 '13 at 12:11
  • @gud I will right what I did, x^2-6x*5=(x-3)^2-4=4[((x-3)/2)^2-1], then x-3/2=tan@, dx=2sec^2@ d@ so the integral will be 1/2(sec^2@/sec^2@)= 1/2 @=1/2 arctan(x-3/2), and now 1/2(lim{t->1-}[arctan(x-3/2) (from 0 to t)]+ lim {t->1+}[arctan(x-3/2) (from t to 3)) please help – user32104 Dec 30 '13 at 12:22
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    @user32104 It seems to me you're using the identity $(\tan(\theta))^2-1=(\sec(\theta))^2$, which is false. Correct would be $(\tan(\theta))^2+1=(\sec(\theta))^2$ – Git Gud Dec 30 '13 at 12:30
  • Omg, thank you finally I know what's wrong – user32104 Dec 30 '13 at 12:36
  • @user32104 No problem. – Git Gud Dec 30 '13 at 12:37
  • @user32104: We would have been able to spot that mistake if you had included your work in the question. – Michael Albanese Dec 30 '13 at 12:39
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enter image description here

This is a plot of the function $y=\frac{1}{x^2-6x+5}$. Does it look integrable to you on the interval $[0,3]$? (Hint: pay special attention to the behaviour of the function around $x=1$.)

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dreamer
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