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 $\operatorname{si}(x)$.

  • $\begingroup$ The question in title is not the same as in the body. Which one is correct ? $\endgroup$ Jan 2, 2022 at 7:25
  • $\begingroup$ @ClaudeLeibovici Are you sure? They look the same to me, but I admittedly suffer from mild dyslexia so I can never be 100% sure haha $\endgroup$
    – David H
    Jan 8, 2022 at 1:04
  • $\begingroup$ @ClaudeLeibovici Nevermind, I see the difference now! But even though their integrands are different functions, it's not too hard to show that the two definite integrals are identical in value for arbitrary $\alpha$. ;) $\endgroup$
    – David H
    Jan 8, 2022 at 1:30

1 Answer 1


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.


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