Evaluate the given integral:

$\int_{0}^{\infty}\cfrac{1}{(x^{2}+1)(x^{2}+9)}dx$

Solution:

(i) $f(z)=\int_{0}^{\infty}\cfrac{1}{(x^{2}+1)(x^{2}+9)}dx$

(ii) $z \cdot f(z)=z \cdot \cfrac{1}{(z^{2}+1)(z^{2}+9)}$

(iii) The poles are $z^{2}+1^{2}=0$ and $z^{2}+3^{2}=0$

$z=\pm i$ and $z=\pm 3i$ lies in the upper half of the z-plane.

(iv) Residue at (z=i)

$\lim _{ z\longrightarrow i }{ (z-i)\cfrac { 1 }{ (z-i)(z+i)({ z }^{ 2 }+{ 3 }^{ 2 }) } } = \lim _{ z\longrightarrow j }{ \cfrac { 1 }{ (z+i)({ z }^{ 2 }+{ 3 }^{ 2 }) } } =\cfrac{1}{(i+i)(i^{2}+3^{2})}$

$=\cfrac{1}{2i(-1+9)} = \cfrac{1}{2i(8)} = \cfrac{1}{16i}$

Similarly, Residue at $z=3i$,

$\lim _{ z\longrightarrow 3i }{ (z-3i)\cfrac { 1 }{ ({ z }^{ 2 }+{ 1 }^{ 2 })(z-3i)(z+3i) } } = \lim _{ z\longrightarrow j }{ \cfrac { 1 }{ (z+3i)({ z }^{ 2 }+1) } } =\cfrac{1}{(3i+3i)(9i^{2}+1)}$

$=\cfrac{1}{-8(6i)} = -\cfrac{1}{48i}$

$f(z)=\int_{0}^{\infty}\cfrac{1}{(x^{2}+1)(x^{2}+9)}dx=2 \pi i \left[ \cfrac{1}{16i} - \cfrac{1}{48i} \right] = \cfrac{2 \pi i}{16} \left[1- \cfrac{1}{3} \right] = \cfrac{ \pi }{8} \left[ \cfrac{3-1}{3} \right] = \pi \left[ \cfrac{2}{24} \right] = \pi \left[ \cfrac{1}{12} \right]$