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Question Number 66464 by mathmax by abdo last updated on 15/Aug/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+3){(x^2+8)}^2}$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+3){(x^2+8)}^2}\Rightarrow 2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x^2+3){(x^2+8)}^2}$$$$let\phi \left(z\right)=\frac{1}{(z^2+3){(z^2+8)}^2}polesof\phi ?$$$$\phi \left(z\right)=\frac{1}{(z-i\sqrt{3})(z+i\sqrt{3}){(z-i2\sqrt{2})}^2{(z+i2\sqrt{2})}^2}thepolesof\phi are$$$$\stackrel{-}{+}i\sqrt{3}and\stackrel{-}{+}2i\sqrt{2}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,i\sqrt{3})+Res(\phi ,2i\sqrt{2})\}$$$$Res(\phi ,i\sqrt{3})={lim}_{z\to i\sqrt{3}}(z-i\sqrt{3})\phi \left(z\right)=\frac{1}{2i\sqrt{3}{(-3+8)}^2}=\frac{1}{50i\sqrt{3}}$$$$Res(\phi ,2i\sqrt{2})={lim}_{z\to 2i\sqrt{2}}\frac{1}{(2-1)!}{\left\{{(z-2i\sqrt{2})}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to 2i\sqrt{2}}{\left\{\frac{1}{(z^2+3){(z+2i\sqrt{2})}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to 2i\sqrt{2}}-\frac{2z{(z+2i\sqrt{2})}^2+2(z+2i\sqrt{2})(z^2+3)}{{(z^2+3)}^2{(z+2i\sqrt{2})}^2}$$$$={lim}_{z\to 2i\sqrt{2}}-\frac{2z(z+2i\sqrt{2})+2(z^2+3)}{{(z^2+3)}^2(z+2i\sqrt{2})}$$$$=-\frac{4i\sqrt{2}\left(4i\sqrt{2}\right)+2(-8+3)}{{(-8+3)}^2\left(4i\sqrt{2}\right)}=-\frac{-32-10}{100i\sqrt{2}}=\frac{42}{100i\sqrt{2}}=\frac{21}{50i\sqrt{2}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{\frac{1}{50i\sqrt{3}}+\frac{21}{50i\sqrt{2}}\}=\frac{\pi }{25\sqrt{3}}+\frac{21\pi }{25\sqrt{2}}=2I$$

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