- #1

Tony1

- 17

- 0

How can we prove $(1)?$

$$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={\pi\over (4a)^3}\tag1$$

$$\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-\int_{0}^{\infty}\mathrm dx{\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}\tag2$$

$$\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}\tag3$$

$${1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}\tag4$$

So far...