Complex Variables and Partial Differential Equations (3130005)

Question-3c

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (c)

Using residue theorem, evaluate $\int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{2}}$ .

I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{1}{2} \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{1}{2} \int_{C}^{} \frac{d z}{{(z^{2} + 1)}^{2}} - - - - - (1)

Where, C : Upper half of | z | = R .

Let,

f (z) = \frac{1}{{(z^{2} + 1)}^{2}} = \frac{1}{{(z + i)}^{2} {(z - i)}^{2}}

\Rightarrow z_{0} = i

is pole of order

n = 2

in

C

.

Res (f (z), z_{0}) = \frac{1}{(n - 1)!} \lim_{z \to z_{0}} \frac{d^{n - 1}}{{dz}^{n - 1}} [{(z - z_{0})}^{n} f (z)]

R e s (f (z), z_{0}) = \frac{1}{1!} \lim_{z \to i} \frac{d}{dz} [{(z - i)}^{2} \frac{1}{{(z + i)}^{2} {(z - i)}^{2}}] = \lim_{z \to i} \frac{d}{dz} [\frac{1}{{(z + i)}^{2}}]

R e s (f (z), z_{0}) = \lim_{z \to i} [- \frac{2}{{(z + i)}^{3}}] = - \frac{2}{{(2 i)}^{3}} = \frac{2}{8 i} = \frac{1}{4 i}

By [ 1 ] and residue theorem, We have

I = \frac{1}{2} [2 πi (sum of residue)]

I = \frac{1}{2} [2 πi (\frac{1}{4 i})]

I = \frac{π}{2}

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