× Close
Join the Ques10 Community
Ques10 is a community of thousands of students, teachers, and academic experts, just like you.
Join them; it only takes a minute
Sign up
Question: Evaluate $ \int_{- \infty}^{\infty} \frac{x^2+x+2}{x^4+10x^2+9} dx $ using Contour Integration.
0

Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 89 views
ADD COMMENTlink
modified 13 days ago  • written 3 months ago by gravatar for Manan Bothra Manan Bothra0
0

$ 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} $

ADD COMMENTlink
written 13 days ago by gravatar for Manan Bothra Manan Bothra0
Please log in to add an answer.


Use of this site constitutes acceptance of our User Agreement and Privacy Policy.