$\int_0^\infty\frac{\cos6x-\cos4x}x\,dx$

Question

I am not able to solve this integral. Please someone help. $$\int_0^\infty\frac{\cos6x-\cos4x}x\,dx$$

Answer 1

Let $a,b,c>0$ $$f(a) = \int_0^\infty\frac{(\cos bx-\cos cx)e^{-ax}}x\,dx$$

By differentiation

$$f'(a) = \int_0^\infty(\cos cx-\cos bx)e^{-ax}\,dx = \frac{a}{c^2+a^2}-\frac{a}{b^2+a^2} $$

Now integrate $$f(a) = \frac{1}{2} \log \left( \frac{c^2+a^2}{b^2+a^2}\right)$$

$$f(0) =\int_0^\infty\frac{(\cos bx-\cos cx)}x\,dx= \log \left(\frac{c}{b}\right)$$