Question

The value of ${\int }_{{}_{{}_0}}^{{}^{{}^{\infty }}}\frac{dx}{{\left(a^2+x^2\right)}^{{}^7}}$ is equal to

• A. $\frac{231}{2047}\left(\frac{1}{a^{13}}\right)$
• B. $\frac{235}{2048}\left(\frac{1}{a^{13}}\right)$
• C. $\frac{232}{2047}\left(\frac{1}{a^{13}}\right)$
• D. $\frac{231}{2048}\left(\frac{1}{a^{13}}\right)$

Answer 1

Answer: (D)

Let I =${\int }_{{}_{{}_0}}^{{}^{{}^{\infty }}}\frac{dx}{{\left(a^2+x^2\right)}^{{}^7}}$
Put $x=atan\theta \Rightarrow dx=asec2\theta d\theta$
limit at x = $0\Rightarrow \theta =0\&x=\infty \Rightarrow \theta =\frac{\pi }{2}$
$\therefore I={\int }_{{}_0}^{{}^{\frac{\pi }{2}}}\frac{ase{c}^2\theta }{a^{14}{\left(1+ta{n}^2\theta \right)}^7}d\theta$
$=\frac{1}{{}_{a}13}{\int }_{{}_0}^{{}^{\frac{\pi }{2}}}\frac{1}{se{c}^{12}\theta }d\theta =\frac{1}{{}_{a}13}.{\int }_{{}_0}^{{}^{\frac{\pi }{2}}}co{s}^{12}\theta .d\theta$
But
${\int }_{{}_0}^{{}^{\frac{\pi }{2}}}si{n}^{2m-1}\theta .d\theta =\frac{1}{2}B\left(m,n\right)$
$\&B\left(m,n\right)=\frac{mn}{m+n}\&\frac{1}{2}=\pi$
$\therefore I={\int }_{{}_0}^{{}^{\frac{\pi }{2}}}si{n}^0\theta .co{s}^{12}\theta d\theta$
$\therefore m=\frac{1}{2},n=\frac{13}{2}$
$\therefore I=\frac{1}{{}_{2a}13}\frac{\frac{1}{2}.\frac{13}{2}}{\frac{1}{2}+\frac{13}{2}}$
$=\frac{1}{{}_{2a}13}.\frac{\pi .\frac{11}{2}.\frac{9}{2}.\frac{7}{2}.\frac{5}{2}.\frac{3}{2}.\frac{1}{2}.\pi }{\mathrm{6.5.4.3.2.1}}$
$=\left(\frac{231}{2048}.\frac{1}{a^{13}}\right)\pi$