Question: Evaluate $\int_{0}^{\infty} \frac{1}{(x^2+1)(x^2+9)} dx$
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 266 views
 modified 12 months ago  • written 16 months ago by
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$f(x) = \frac{1}{(x^2+1)(x^2+9)} \\ f(z) = \frac{1}{(z^2+1)(z^+9)}$

Consider,

$\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+9)} \\ = \int_c \frac{1}{(z^2+1)(z^+9)}$

Consider, $f(z) = \frac{1}{(z^2+1)(z^+9)}$

For poles, $z^2+1=0 \hspace{0.20cm} \& \hspace{0.20cm} z^9 + 9 = 0 \\ \therefore z = \pm i \hspace{0.20cm} \& \hspace{0.20cm} z = \pm 3i$

$\therefore z = \pm i, \pm 3i$ are the four poles of which we consider only poles lying in the upper half of the circle |z| = R

$\therefore z = i,3i$ are the only poles which lie inside the circle.

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

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

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

$lim_{z \to 3i} (z-3i)[\frac{1}{(z^+1)(z-3i)(z+3i)}] \\ = \frac{1}{(3i)^2 + 1][3i+3i]} = \frac{1}{(-9+1)(6i)} = \frac{1}{-8(6i)} = -\frac{1}{8(6i)} = -\frac{1}{48i}$

Sum of residues $= \frac{1}{16i} - \frac{1}{48i} = \frac{3}{48i} - \frac{1}{48i} = \frac{2}{48i} = \frac{1}{24i}$

$\therefore \int_c \frac{1}{(z^2+1)(z^2+9)} \,\, dz = 2 \pi i[\frac{1}{24i}] = \frac{\pi}{12}$

$\int_{-\infty}^{\infty} \frac{1}{(x^2+1)(x^2+9)} \,\, dx = \frac{\pi}{12} \\ 2 \int_0^{\infty} \frac{1}{(x^2+1)(x^2+9)} \,\, dx = \frac{\pi}{12} \\ \int_0^{\infty} \frac{1}{(x^2+1)(x^2+9)} \,\, dx = \frac{\pi}{24}$