# Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$

Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$

Any solutions or hints are greatly appreciated.

I know I can rewrite the integral as $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ but I'm not sure how to proceed.

• partial fraction decomposition is made for that : transforming a rational function into a sum of easily integrable functions. the only difficulty is that it involves factorizing the denominator. – reuns Apr 21 '16 at 19:22
• Since January 20 all but one of the questions you have posted have lacked context. Several of your posts have been closed for this reason, yet you continue to ignore all warnings that the posting of questions without any explanation of your own thoughts or effort to solve them is not allowed on this site. – heropup Apr 21 '16 at 20:21

Well, $\int_{-\infty}^{+\infty}\frac{-3x}{x^4+10x^2+9}\,dx = 0$, hence the problem boils down to computing: $$\int_{-\infty}^{+\infty}\frac{(x^2+2)}{(x^2+1)(x^2+9)}\,dx=\frac{1}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\,dx+\frac{7}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+9}$$ that is trivially equal to $\left(\frac{1}{8}+\frac{7}{24}\right)\pi = \color{red}{\large\frac{5\pi}{12}}$.

Footnote: how to find the coefficients $\frac{1}{8}$ and $\frac{7}{8}$ very fast. We know that for some $A,B$ $$g(z)=\frac{z+2}{(z+1)(z+9)}=\frac{A}{z+1}+\frac{B}{z+9}$$ must hold. On the other hand, $A=\lim_{z\to -1}g(z)(z+1)$ as well as $B=\lim_{z\to -9}g(z)(z+9)$, so: $$A = \lim_{z\to -1}\frac{z+2}{z+9},\qquad B=\lim_{z\to -9}\frac{z+2}{z+1}.$$

• I applaud the use of odd symmetry of the "middle term" in the integrand, somewhat it is often helpful to be alert for. – colormegone Apr 21 '16 at 20:01

You can simply use partial fractions $${{x^2-3x+2}\over {x^4+10x^2+9}} = \frac{1-3x}{8(x^2+1)}+\frac{3x+7}{8(x^2+9)},$$ which will allow you to find the antiderivative.

• Even easier if you notice that we may get rid of the $-3x$. – Jack D'Aurizio Apr 21 '16 at 19:46
• Well seen yes ! – C. Dubussy Apr 21 '16 at 19:49

Split the integral into two integrals, such that you have only one "bad" point. Now:

$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx= \int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx + \int_{0}^{\infty}{{x^2-3x+2}\over {x^4+10x^2+9}}dx$$

Now deal with the integrals separately:

$$\int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \int_{t}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_t^0 = \frac{9}{48}\ln(9) + \frac{10\pi}{48}$$

$$\int_{0}^{\infty} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \int_{0}^{t} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_0^t = -\frac{9}{48}\ln(9) + \frac{10\pi}{48}$$

Summing them you will get:

$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \frac{9}{48}\ln(9) + \frac{10\pi}{48} -\frac{9}{48}\ln(9) + \frac{10\pi}{48} = \frac{5\pi}{12}$$

NOTE: I skipped the calculation of the integrals, but it can be easily done by partial fraction decomposition and it will be reduced to something more familiar.

Transform your integral into a contour integral that travels from -R to R, then follows a semicircle to -R. As R -> $\infty$, the arc of this contour integral goes to 0, as seen by the ML inequality, and the line goes to the real integral. The function has poles at $\pm i,\pm 3i$, of which only $i,3i$ is inside the contour. Evaluating the residues at those points, you get the answer.