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How to calculate the principal part of this improper integral via contour integration?

\begin{equation} P\int_{0}^{+\infty}\frac{dx}{x^2+x-2} \end{equation}

I have seen some examples where you integrate along a semicircle and then take the limit $R\to\infty$, where $R$ is the radius. But here the integrand is not a even function and in addition there are poles on the real line (that's why is divergent)... Any help is appreciated.

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    I simply meant that a problem can be solved in different ways, following different lines of thought and you don't have to be single-minded about a particular way to tackle it if it doesn't work... anyway, I agree that in an exercise like this nobody cares if the final digit is wrong or not...2012-02-18

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Let's factorize the denominator : $x^2+x-2=(x+2)(x-1)$

so that $\displaystyle \frac1{x^2+x-2}=\frac13\left(\frac1{x-1}-\frac1{x+2}\right)$

Let's try a direct proof without complex integration. $P.V. \int_0^{\infty} \frac 1{x^2+x-2}=\frac13 P.V.\int_0^{\infty} \frac1{x-1}-\frac1{x+2} dx=$ $ =\frac13 \lim_{\epsilon\to 0}\left[\int_0^{1-\epsilon} \frac1{x-1}-\frac1{x+2} dx+\int_{1+\epsilon}^{\infty} \frac1{x-1}-\frac1{x+2} dx\right] $

$ =\frac13 \lim_{\epsilon\to 0}\left[ \left[\log(1-x)-\log(x+2)\right]_0^{1-\epsilon}+ \left[\log(x-1)-\log(x+2)\right]_{1+\epsilon}^{\infty}\right] $ $ =\frac13 \lim_{\epsilon\to 0}\left[\log(\epsilon)-\log(3-\epsilon)+\log(2)-\log(\epsilon)+\log(3+\epsilon)\right] $ (using $\lim_{R\to \infty}\log\left(\frac{R-1}{R+2}\right)=0$) $ =\frac{\log(2)}3 $

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    More details about contour integration with poles on the contour may be found in this [answer](http://math.stackexchange.com/questions/478534/two-questions-regarding-mathrm-li-from-edwards/479237#479237).2014-09-20