Question: Evaluate $\int_{- \infty}^{\infty} \frac{x^2+x+2}{x^4+10x^2+9} dx$ using Contour Integration.
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 354 views
 modified 12 months ago  • written 16 months ago by
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$f(x) = \frac{x^2+x+2}{x^4+10x^2+9} \\ f(z) = \frac{z^2+z+2}{z^4+10z^2+9}$

For poles $z^4 + 10z^2 + 9 = 0$

$z^4 + 9z^2 + z^2 + 9 = 0 \\ z^2(z^9+9) + 1(z^2+9) = 0 \\ (z^2+1)(z^2+9) = 0 \\ \therefore z^2 = -1 \hspace{0.20cm} \& \hspace{0.20cm} z^2 = -9 \\ z = \pm i \hspace{0.20cm} \& \hspace{0.20cm} z = \pm 3i$

$\therefore z = i,-i,3i,-3i$ are the four poles of which z = i and z = 3i are the only two poles which lie inside the circle.

Residue of f(z) at z = i is:

$\lim_{z \to i} (z-i)[ \frac{z^2+z+2}{z^4+10z^2+9}] \\ = \lim_{z \to i} (z-i)[\frac{z^2+z+2}{(z^2+9)(z^2+1)}] \\ = \lim_{z \to i} (z-i)[\frac{z^2+z+2}{(z^2+9)(z+i)(z-i)}] \\ = \frac{i^2+i+2}{(i^2+9)(i+i)} = \frac{-1+i+2}{(-1+9)(2i)} = \frac{1+i}{(8)(2i)} = \frac{1+i}{16i}$

Residue of f(z) at z = 3i is:

$\lim_{z \to 3i} (z-3i)[\frac{z^2+z+2}{(z^2+9)(z^2+1)}] \\ = \lim_{z \to 3i} (z-3i)[\frac{z^2+z+2}{(z-3i)(z+3i)(z^2+1)}] \\ = \frac{(3i)^2+3i+2}{(3i+3i)[(3i)^2+1]} = \frac{-9+3i+2}{6i(-9+1)} = \frac{-7+3i}{-48i} = \frac{7-3i}{48i}$

Sum of residues $= \frac{3(1+i)}{3(16i)} + \frac{7-3i}{48i}$

$\frac{3+3i37-3i}{48i} = \frac{10}{48i} = \frac{5}{24i} \\ \therefore \int_c \frac{z^2+z+2}{z^4+10z^2+9} \,\, dz = 2 \pi i[\frac{5}{24i}] = \frac{5 \pi}{12} \\ \therefore \int_{- \infty}^{\infty} \frac{x^2+x+2}{x^4+10x^2+9} \,\, dx = \frac{5\pi}{12}$