# Write $\int\limits_0^1\cos(2x)\frac{\operatorname{Ci}(2-2x)-\operatorname{Ci}(2+2x)}{x}dx$ as a series

While I was trying calculate $$\int_0^1\int_0^1\frac{\sin(x-y)\sin(x+y)}{x^2-y^2}dxdy,$$ I am wondering

Question. Is it possible to write $$\int_0^1\cos(2x)\frac{\operatorname{Ci}(2-2x)-\operatorname{Ci}(2+2x)}{x}dx\tag{1}$$ as a series? Here with $\operatorname{Ci}(x)$ we are denoting the cosine integral, see the definition, if you don't know it, in this MathWorld. Thanks in advance.

I try to continue to compute it, but now I don't know how get an expression of $(1)$ as a series. I know that is required to combine with the series expansion of the cosine integral and with Newton's Binomial theorem.

• Hint to compute your integral, $x^2-y^2=(x+y)(x-y)$ – FDP Jul 9 '17 at 19:29
• To get a series expansion perform integration by parts, an antiderivative for $1/x$ is $\ln x$ – FDP Jul 9 '17 at 19:32
• Many thanks for your notes. I take those in my notebook and I am going to thinks in these @FDP – user243301 Jul 9 '17 at 19:55 