Evaluate $\int_{0}^{\pi/2}\left [\text{ Ci}(\alpha\sin\theta)^2+\left(\frac{\pi}{2}-\text{ Si}(\alpha\sin\theta)\right)^2\right ]\text{d}\theta$ There is a well-known identity for a complex number such that $\alpha\ne0$ (by differentiating both sides):
$$\int_{0}^{\frac{\pi}{2} }\operatorname{Ci}(\alpha \cos\theta)\text{d}\theta
=  \frac{\pi}{2} \left [ \gamma+\ln\left ( \frac{\alpha}{2}  \right )  \right ]
-\frac{\pi}{16}\alpha^2 {}_2F_3
\left ( 1,1;2,2,2;-\frac{\alpha^2}{4}  \right ) .$$
Where $\operatorname{Ci}(x)=-\int_{x}^{\infty} \frac{\cos(t)}{t}\text{d}t$, ${}_2F_3$ is Generalized Hypergeometric function and $\gamma=0.5772156649015932...$ is Euler-Mascheroni constant.
But how can we evaluate this integral?
$$\int_{0}^{\frac{\pi}{2} } 
\left [ \operatorname{Ci}
(\alpha\cos\theta)^2+\left ( \frac{\pi}{2}-\operatorname{Si}(\alpha\sin\theta) \right ) ^2 \right ]\mathrm{d}\theta$$
with $\operatorname{Si}(x)=\int_{0}^{x} \frac{\sin(t)}{t}\text{d}t.$
Furthermore, $\frac{\pi}{2}-\operatorname{Si}(x)$ can be denoted as $\operatorname{si}(x)$.
 A: I am skeptical about a possible closed form.
What I suppose is that we could do is to use the infinite series
$$\text{Ci}(x)=\gamma+\log(x)+\sum_{n=1}^\infty(-1)^n \frac{ x^{2 n}}{2 n (2 n)!}$$
$$\text{Si}(x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2 n-1}}{(2 n-1) (2 n-1)!}$$ and make $x=\alpha\, \text{something}$ (as I commented, what you wrote in title is not the same as in the body of the question).
This will give
$$\int_{0}^{\frac{\pi}{2} }\text{Ci}(\alpha \cos\theta)\,d\theta=\frac{1}{2} \pi  \left(\gamma+\log \left(\frac{\alpha }{2}\right) \right)+\frac{\sqrt{\pi }}4\sum_{n=1}^\infty(-1)^n  \frac{\alpha^{2n} \,\Gamma \left(n+\frac{1}{2}\right)}{ n \,n! \,(2 n)! }$$
$$\frac{\sqrt{\pi }}4\sum_{n=1}^\infty(-1)^n  \frac{\alpha^{2n} \,\Gamma \left(n+\frac{1}{2}\right)}{ n \, n! \,(2 n)! }=-\frac{\pi}{16}   \alpha^2 \, _2F_3\left(1,1;2,2,2;-\frac{\alpha^2}{4}\right)$$ which is your first identity.
In the same manner
$$\int_{0}^{\frac{\pi}{2} }\text{Si}(\alpha \sin\theta)\,d\theta=\frac{\sqrt{\pi }}2\sum_{n=1}^\infty(-1)^{n -1} \frac{\alpha ^{2 n-1} \Gamma (n)}{(2 n-1) (2 n-1)! \Gamma \left(n+\frac{1}{2}\right)}$$
$$\int_{0}^{\frac{\pi}{2} }\text{Si}(\alpha \sin\theta)\,d\theta=\alpha  \, _2F_3\left(\frac{1}{2},1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\frac{\alpha
   ^2}{4}\right)=\int_{0}^{\frac{\pi}{2} }\text{Si}(\alpha \cos\theta)\,d\theta$$
Now, where comes the problem is with the squaring of the series. If we expand the square of the first integral, we have
$$\Big[\text{Ci}(\alpha \cos\theta)\Big]^2=\Sigma ^2+2 \gamma  \Sigma +2 \Sigma  \log (x)+\log ^2(x)+2 \gamma  \log (x)+\gamma ^2$$ and the integration of the first term makes problem. For the third term which looks simpler
$$\int_{0}^{\frac{\pi}{2} }\frac{ \alpha^{2 n} \cos ^{2 n}(\theta ) \log (\alpha  \cos (\theta ))}{2 n (2 n)!}\,d\theta=\frac{\sqrt{\pi } \alpha^{2 n} \Gamma \left(n+\frac{1}{2}\right) \left(2 \log (\alpha
   )+H_{n-\frac{1}{2}}-H_n\right)}{8 n (2 n)! \Gamma (n+1)}$$ and even the partial summation does not seem to lead to anything.
