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i know that this is an improper integral, but when you evaluate the limits as $x\to (3/2)^-$ and $x\to (3/2)^+$, you get positive and negative infinity but I am not sure if you can cancel them out...any help is very appreciated!

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No, you cannot cancel them out, as each improper partial integral must converge, yet

$$\int_0^{\frac32-\epsilon}\frac x{3-2x}dx=\frac12\int_0^{\frac32-\epsilon}\left(\frac3{3-2x}-1\right)dx=\left.-\frac34\log(3-2x)\right|_0^{\frac32-\epsilon}-\frac12\left(\frac32-\epsilon\right)$$

yet the limit

$$\lim_{\epsilon\to 0^+}\log2\epsilon\;\;\;\text{isn't finite}$$

and this is enough to conclude the integral diverges.

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As Timbuc answered, the integral doesn't exist in the Riemann sense. It's not the limit of a Riemann sum.

However, there is a way to deal with these integrals, called the Cauchy principal value integral. Introductory Calculus textbooks don't address this, but if you're curious:

After polynomial division, you get:

$$-\frac 3 2 \int_0^2 \frac 1 {3-2x} \, \mathrm dx - \int_0^2 \frac 1 2 \, \mathrm dx$$

The second integral doesn't cause us any problems, so I won't deal with it. Timbuc's answer explains why the Riemann integral doesn't exist for the first term. Here's a different type of integral, called the Cauchy Principal Value integral:

$$-\frac {3}2 \, PV \int_0^2 \frac 1 {3-2x} \, \mathrm dx $$

$$ = - \frac 3 2 \lim_{\epsilon \to 0^+} \left({\int_0^{-3/2 - \epsilon} \frac 1 {3-2x} \, \mathrm dx + \int_{-3/2 + \epsilon}^2 \frac 1 {3-2x} \, \mathrm dx }\right)$$

$$= - \frac 3 4 \lim_{\epsilon \to 0^+}\left({\ln |-2 \epsilon|- \ln 3 + \ln |-1| - \ln (2\epsilon) }\right) $$

$$= - \frac 3 4 \lim_{\epsilon \to 0^+}\left({\ln |-2\epsilon|- \ln \left({2 \epsilon}\right)}\right)- \frac 3 4 \lim_{\epsilon \to 0^+}\left({\ln |-1|- \ln 3}\right)$$

$$= \frac 3 4 \ln 3 $$

I emphasize that this is not a Riemann integral. That's what the $PV$ is for.

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  • $\begingroup$ Good point and good answer. +1 $\endgroup$ – Timbuc Feb 27 '15 at 14:22

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