How do i find $f(0,p)$ for $$f(x,p) = \int \frac{\sin(2x)}{x} C_i(x+p) dx$$
$C_i(x)$ is Cosine Integral. Obviously, if I could solve the integral, I would have substituted $x$ with $0$. But I don't know how to solve it. I wonder if there is a general technique for finding the value of this function at a particular point, even if I can't solve it generally for all values of $x$. Ignore the constant of integration (the usual +C term for indefinite integrals)
What I would probably try is to expand the cosine integral around $x=0$ $$\text{Ci}(x+p)=\sum_{n=0}^\infty \frac {a_n}{n!\,p^n}x^n$$ and face the integrals $$I_n(x)=\int x^{n-1} \sin (2 x)\,dx=\frac{1}{2} i x^n (E_{1-n}(-2 i x)-E_{1-n}(2 i x))$$ that is to say $$I_n(0)=-2^{-n} \sin \left(\frac{\pi n}{2}\right) \Gamma (n,0)$$ So, $$I_{2n}(0)=0 \qquad \text{and}\qquad I_{2n+1}(0)=(-1)^{n+1}2^{-(2 n+1)} \Gamma (2 n+1,0)$$
This would give $$f(0,p)= \sum_{n=0}^\infty (-1)^{n+1}\frac{ \Gamma (2 n+1,0)}{(2p)^{2n+1} \,(2n+1)! } a_{2n+1}$$
Now $$a_{2n+1}=P_{n}\, \cos(p)+Q_{n} \,\sin(p)$$ and the first polynomials are $$\left( \begin{array}{cc} n & P_n \\ 0 & 1 \\ 1 & -p^2 +2\\ 2 & p^4-12 p^2+24 \\ 3 & -p^6+30 p^4-360 p^2+720 \\ 4 & p^8-56 p^6+1680 p^4-20160 p^2+40320 \\ 5 & -p^{10}+90 p^8-5040 p^6+151200 p^4-1814400 p^2+3628800 \\ 6 & p^{12}-132 p^{10}+11880 p^8-665280 p^6+19958400 p^4-239500800 p^2+479001600 \end{array} \right)$$
$$\left( \begin{array}{cc} n & Q_n \\ 0 & 0 \\ 1 & 2 p \\ 2 & -4 p^3+24p \\ 3 & 6 p^5-120 p^3+720 p \\ 4 & -8 p^7+336 p^5-6720 p^3+40320 p \\ 5 & 10 p^9-720 p^7+30240 p^5-604800 p^3+3628800 p \\ 6 & -12 p^{11}+1320 p^9-95040 p^7+3991680 p^5-79833600 p^3+479001600 p \end{array} \right)$$ where interesting patterns appear.
