An integral with a parameter.

The problem is to evaluate (with proof) $$f(a) = \int_{0}^{\infty} \frac{x^2}{(x^2 - a^2)^2 + x^2} \text{ d}x$$ where $a$ is a real number.


Scroll down for a solution.






Surprisingly, the answer is independent of $a$!.

For $a=0$, the integral is the standard $\int_{0}^{\infty} \frac{dx}{1+x^2} = \frac{\pi}{2}$. A valuable tool when evaluating integrals is substitution and it really helps here. First, assume $a \neq 0$, and make the substitution $u = \frac{a^2}{x}$ This gives us that $$f(a) = \int_{0}^{\infty} \frac{\frac{a^4}{u^2}}{((\frac{a^4}{u^2} - a^2)^2 + \frac{a^4}{u^2})} \frac{a^2\text{d}u}{u^2}$$ $$= \int_{0}^{\infty} \frac{a^2}{(u^2 - a^2)^2 + u^2} \text{ d}u$$ $$= \int_{0}^{\infty} \frac{a^2}{(x^2 - a^2)^2 + x^2} \text{ d}x$$ Thus by adding the original, we get that $$2f(a) = \int_{0}^{\infty} \frac{x^2 + a^2}{(x^2 - a^2)^2 + x^2} \text{ d}x$$ $$= \int_{0}^{\infty} \frac{1 + \frac{a^2}{x^2}}{(x - \frac{a^2}{x})^2 + 1} \text{ d}x$$ Now make the substitution $u = x - \frac{a^2}{x}$ to get $$2f(a) = \int_{-\infty}^{\infty} \frac{1}{u^2 +1} \text{ d}u = \pi$$ And so $$\int_{0}^{\infty} \frac{x^2}{(x^2 - a^2)^2 + x^2} \text{ d}x = \frac{\pi}{2}$$ Note: The substitutions need to be justified because technically, the integral is an improper integral, but we leave that to the reader :)