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Find $$\int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)}$$


My Attempt:

Let $$\begin{align}\frac{1}{(n-2)(3-n)}&=\frac{A}{n-2}+\frac{B}{3-n} \\ &= \frac{A(3-n)+B(n-2)}{(n-2)(3-n)}\\ \Rightarrow 1 &= A(3-n)+B(n-2) \\ &= 3A - An+Bn-2B \\ &= n(B-A)+3A-2B.\end{align}$$ Equating coefficients on both sides, we obtain $$B-A=0 \qquad and \qquad 3A-2B=1.$$ $$\therefore A=B=1.$$ $$$$ $$\begin{align}\therefore \int\limits_{2}^{3}\frac{\mathrm dn}{(n-2)(3-n)} &= \int\limits_{2}^{3}\bigg(\frac{1}{n-2}+\frac{1}{3-n}\bigg)\,\mathrm dn \\ &= \int\limits_{2}^{3}\frac{1}{n-2}\,\mathrm dn+\int\limits_{2}^{3}\frac{1}{3-n}\,\mathrm dn \\ &=\left[\log(n-2)\right]_{\small 2}^{\small 3}+\left[\log(3-n)\right]_{\small 2}^{\small 3} \\ &= \left[\log(3-2)-\log(2-2)\right]+\left[\log(3-3)+\log(3-2)\right] \\ &= \left[\log 1-\log 0\right]+\left[\log 0 - \log 1\right]\end{align}$$ but $\log 0$ is undefined, thus my answer is coming undefined. Am I doing something wrong in the solution?


Thank you in advance.

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  • $\begingroup$ Actually this integral is divergent. $x=2$ and $x=3$ are two vertical asymptotes of the given integrand. $\endgroup$
    – little o
    Commented Apr 14, 2019 at 7:01

2 Answers 2

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No.

There are two singularities at $n = 2$ and $n = 3$, as can be determined by inspecting the denominator, and they are quite strong. The integral does not converge.

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  • $\begingroup$ Whats quite strong mean $\endgroup$ Commented Apr 14, 2019 at 7:09
  • $\begingroup$ @MikeySpivak - Basically, how quickly the function goes to infinity as the singularity is approached. Generally speaking, the larger the negative exponent of a negative-degree term, the stronger the singularity is. $\endgroup$ Commented Apr 14, 2019 at 7:12
  • $\begingroup$ More quantitatively, a singular term $x^{-\alpha}$ with $\alpha \ge 1$ is strong enough to diverge when integrated (with respect to $x$) up to its singular point, but with $0 \le \alpha < 1$, it actually is weak enough that it can converge. $\endgroup$ Commented Apr 14, 2019 at 7:14
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No, there is nothing wrong. You have\begin{align}\int_2^3\frac1{(x-2)(x-3)}\,\mathrm dx&=\int_2^{5/2}\frac1{(x-2)(x-3)}\,\mathrm dx+\int_{5/2}^3\frac1{(x-2)(x-3)}\,\mathrm dx\\&=\lim_{t\to2^+}\int_t^{5/2}\frac1{(x-2)(x-3)}\,\mathrm dx+\lim_{t\to3^-}\int_{5/2}^t\frac1{(x-2)(x-3)}\,\mathrm dx.\end{align}None of these limits exist, since the function that's being integrated behaves as $\frac1{x-3}$ near $3$ and as $\frac1{x-2}$ near $2$.

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