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

Question

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.

Answer 1

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 $\left(\frac{1}{8}+\frac{7}{24}\right)\pi = \color{red}{\large\frac{5\pi}{12}}$.

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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: $$ A = \lim_{z\to -1}\frac{z+2}{z+9},\qquad B=\lim_{z\to -9}\frac{z+2}{z+1}.$$

Answer 2

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.

Answer 3

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.

Answer 4

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.