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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.

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  • $\begingroup$ Hint to compute your integral, $x^2-y^2=(x+y)(x-y)$ $\endgroup$ – FDP Jul 9 '17 at 19:29
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    $\begingroup$ To get a series expansion perform integration by parts, an antiderivative for $1/x$ is $\ln x$ $\endgroup$ – FDP Jul 9 '17 at 19:32
  • $\begingroup$ Many thanks for your notes. I take those in my notebook and I am going to thinks in these @FDP $\endgroup$ – user243301 Jul 9 '17 at 19:55
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I don't know if further simplification is possible.

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  • $\begingroup$ Many thanks for your attention and contribution. You are generous with your nice answers in this site MSE. I am going to study your nice answer. $\endgroup$ – user243301 Jul 15 '17 at 19:28
  • $\begingroup$ Don't mention it. Note : Some typos in my first answer have been corrected. $\endgroup$ – JJacquelin Jul 16 '17 at 5:52

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