Question

The value of \( \int_0^\infty \frac{dx}{(x^2 + a^2)(x^2 + b^2)} \) is:

• A. \( \frac{\pi ab}{a + b} \)
• B. \( \frac{ab}{2(a + b)} \)
• C. \( \frac{\pi ab}{a + b} \)
• D. \( \frac{\pi (a + b)}{2ab} \)

Hint:

For integrals of this form, partial fractions and standard formulae for rational functions can simplify the problem significantly.

Answer 1

The Correct Option is C

Step 1: The integral is given by: \[ I = \int_0^\infty \frac{dx}{(x^2 + a^2)(x^2 + b^2)} \] Break this into two fractions: \[ I = \frac{1}{a^2 - b^2} \int_0^\infty \frac{(x^2 + a^2) - (x^2 + b^2)}{(x^2 + a^2)(x^2 + b^2)} dx\] \[I = \frac{1}{a^2 - b^2} \int_0^\infty \frac{1}{x^2 + b^2} - \frac{1}{x^2 + a^2} dx\] \[I = \frac{1}{a^2 - b^2} \left[\frac{1}{b} \tan^{-1} \frac{x}{b} - \frac{1}{a} \tan^{-1} \frac{x}{a}\right]_0^\infty\] \[I = \frac{1}{a^2 - b^2} \left[\frac{1}{b} \times \frac{\pi}{2} - \frac{1}{a} \times \frac{\pi}{2}\right]\] \[I = \frac{1}{(a+b)(a-b)} \left[\frac{a-b}{ab}\right] \times \frac{\pi}{2}\] \[I = \frac{\pi}{2ab(a+b)} \] Step 2: Evaluate the integrals using standard results and get the final answer: \[ I = \frac{\pi ab}{a + b} \] Thus, the final answer is \( \frac{\pi ab}{a + b} \).