Calculate $\int_1^\infty \frac{ (1-x+[x])(x^{1-\sigma}-x^\sigma)cos(ln x)}{x^2}dx$

Question

Calculate $\int_1^\infty \frac{ (1-x+[x])(x^{1-\sigma}-x^\sigma)cos(ln x)}{x^2}dx $ where [.] denotes the greatest integer function and $0<\sigma<1$.

My try -

$\int_1^\infty \frac{ (1-x+[x])(x^{1-\sigma}-x^\sigma)cos(ln x)}{x^2}dx = \sum_{n=1}^\infty \int_n^{n+1} \frac{ (1-x+n)(x^{1-\sigma}-x^\sigma)cos(ln x)}{x^2} dx Substitute \ ln x=u \Rightarrow x=e^u \Rightarrow dx=e^u du \sum_{n=1}^\infty \int_{ln n}^{ln(n+1)} \frac{ (1-e^u+n)(e^{(1-\sigma)u}-e^{\sigma u})cos(u)}{e^u} dx $ After this i think that the formula for $\int e^{ax} cos bx$ should be applied. But i am stuck .

Answer 1

$\textbf{Hint:}$ Since $\cos\log x = \operatorname{Re}(x^i)$ we can instead do the integral

$$\int_1^\infty (1-\{x\})(x^{1-\sigma}-x^{\sigma})x^{i-2}\:dx$$

Can you take it from here?