The value of $\int_{0}^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)^{7}}$ is equal to

• A

  $\frac{231}{2047}\left(\frac{1}{a^{13}}\right)$
• B

  $\frac{235}{2048}\left(\frac{1}{\mathrm{a}^{13}}\right)$
• C

  $\frac{232}{2047}\left(\frac{1}{\mathrm{a}^{13}}\right)$
• Correct answer
  D

  $\frac{231}{2048}\left(\frac{1}{\mathrm{a}^{13}}\right)$

Solution

Let $I=\int_{0}^{\infty} \frac{d x}{\left(a^{2}+x^{2}\right)^{7}}$ Put $\mathrm{x}=\mathrm{a} \tan \theta \Rightarrow \mathrm{dx}=\mathrm{a} \sec ^{2} \theta \mathrm{d} \theta$ limit at $x=0 \Rightarrow \theta=0 \& x=\infty \Rightarrow \theta=\frac{\pi}{2}$ $\therefore I=\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta}{a^{14}\left(1+\tan ^{2} \theta\right)^{7}} d \theta$ $=\frac{1}{a^{13}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sec ^{12} \theta} d \theta=\frac{1}{a^{13}} \cdot \int_{0}^{\frac{\pi}{2}} \cos ^{12} \theta \cdot d \theta$ But $\int_{0}^{\frac{\pi}{2}} \sin ^{2 m-1} \theta \cdot \cos ^{2 n-1} \theta \cdot \mathrm{d} \theta=\frac{1}{2} \mathrm{~B}(\mathrm{~m}, \mathrm{n})$ $\& \mathrm{~B}(\mathrm{~m}, \mathrm{n})=$
$\therefore \mathrm{I}=\frac{1}{\mathrm{a}^{13}} \int_{0}^{\frac{\pi}{2}} \sin ^{0} \theta \cdot \cos ^{12} \theta \mathrm{d} \theta$ $\therefore \mathrm{m}=\frac{1}{2}, \mathrm{n}=\frac{13}{2}$
$=\left(\frac{231}{2048} \cdot \frac{1}{a^{13}}\right) \pi$

Hence, the correct answer is option: 'D'.