Closed form evaluation of a trigonometric integral in terms of polylogarithms

Question

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Define the function $\mathcal{K}:\mathbb{R}\times\mathbb{R}\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{K}{\left(a,b,\psi,\omega\right)}:=\int_{\psi}^{\omega}\mathrm{d}\varphi\,\arctan{\left(a+b\tan{\left(\varphi\right)}\right)}.\tag{1}$$

It's not too difficult to find a closed-form expression for the integral in the following special case: for all $b\in\mathbb{R}_{>0}\land\omega\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$,

$$\begin{align} \mathcal{K}{\left(0,b,0,\omega\right)} &=\int_{0}^{\omega}\mathrm{d}\varphi\,\arctan{\left(b\tan{\left(\varphi\right)}\right)}\\ &=\frac12\omega^{2}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-b}{1+b}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-b}{1+b},\pi-2\omega\right)},\tag{2}\\ \end{align}$$

where

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.\tag{3}$$

I would like to consider, however, the general evaluation of $\mathcal{K}{\left(a,b,\psi,\omega\right)}$ under the conditions

$$0<a\land\sqrt{1+a^{2}}<b\land-\frac{\pi}{4}\le\psi<\omega\le\frac{\pi}{4}.\tag{4}$$

It is in fact possible to obtain a closed-form for $\mathcal{K}$ in terms of the two-variable dilogarithm in this case as well. But how best to obtain/verify such an expression?

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Answer 1

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The following approach is modeled after Lewin's Polylogarithms and Associated Functions. It's a little cumbersome though, and it seems to break down unless the conditions specified in $(4)$ are met. A more general method would be welcomed.

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Suppose $A\in\mathbb{R}\land B\in\mathbb{R}_{\neq0}\land\phi\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\land\varphi\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Then, $-\frac{\pi}{2}<\phi+\varphi<\frac{\pi}{2}\land-1<\tan{\left(\phi\right)}\tan{\left(\varphi\right)}<1$, and by the tangent angle addition formula we have

$$\tan{\left(\phi+\varphi\right)}=\frac{\tan{\left(\phi\right)}+\tan{\left(\varphi\right)}}{1-\tan{\left(\phi\right)}\tan{\left(\varphi\right)}}.$$

Recall the arctangent addition formula:

$$\arctan{\left(z\right)}+\arctan{\left(w\right)}=\arctan{\left(\frac{z+w}{1-zw}\right)};~~~\small{\left(z,w\right)\in\mathbb{R}^{2}\land zw<1}.$$

Taking $z=A\land w=B\tan{\left(\phi+\varphi\right)}$, we then have

$$\arctan{\left(A\right)}+\arctan{\left(B\tan{\left(\phi+\varphi\right)}\right)}=\arctan{\left(\frac{A+B\tan{\left(\phi+\varphi\right)}}{1-AB\tan{\left(\phi+\varphi\right)}}\right)},$$

provided that $AB\tan{\left(\phi+\varphi\right)}<1$.

Adding the constraint $AB=-\tan{\left(\phi\right)}$, we find that

$$\begin{align} 1-AB\tan{\left(\phi+\varphi\right)} &=1+\tan{\left(\phi\right)}\tan{\left(\phi+\varphi\right)}\\ &=1+\tan{\left(\phi\right)}\frac{\tan{\left(\phi\right)}+\tan{\left(\varphi\right)}}{1-\tan{\left(\phi\right)}\tan{\left(\varphi\right)}}\\ &=\frac{1+\tan^{2}{\left(\phi\right)}}{1-\tan{\left(\phi\right)}\tan{\left(\varphi\right)}}\\ &=\frac{\sec^{2}{\left(\phi\right)}}{1-\tan{\left(\phi\right)}\tan{\left(\varphi\right)}}>0,\\ \end{align}$$

and so

$$-\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\arctan{\left(B\tan{\left(\phi+\varphi\right)}\right)}=\arctan{\left(\frac{A+B\tan{\left(\phi+\varphi\right)}}{1-AB\tan{\left(\phi+\varphi\right)}}\right)},$$

where

$$\begin{align} \frac{A+B\tan{\left(\phi+\varphi\right)}}{1-AB\tan{\left(\phi+\varphi\right)}} &=\frac{-B^{-1}\tan{\left(\phi\right)}+B\tan{\left(\varphi+\phi\right)}}{1+\tan{\left(\phi\right)}\tan{\left(\varphi+\phi\right)}}\\ &=\frac{-B^{-1}\tan{\left(\phi\right)}+B\left[\frac{\tan{\left(\varphi\right)}+\tan{\left(\phi\right)}}{1-\tan{\left(\varphi\right)}\tan{\left(\phi\right)}}\right]}{1+\tan{\left(\phi\right)}\left[\frac{\tan{\left(\varphi\right)}+\tan{\left(\phi\right)}}{1-\tan{\left(\varphi\right)}\tan{\left(\phi\right)}}\right]}\\ &=\frac{-B^{-1}\tan{\left(\phi\right)}\left[1-\tan{\left(\varphi\right)}\tan{\left(\phi\right)}\right]+B\left[\tan{\left(\varphi\right)}+\tan{\left(\phi\right)}\right]}{\left[1-\tan{\left(\varphi\right)}\tan{\left(\phi\right)}\right]+\tan{\left(\phi\right)}\left[\tan{\left(\varphi\right)}+\tan{\left(\phi\right)}\right]}\\ &=\frac{-\tan{\left(\phi\right)}+\tan^{2}{\left(\phi\right)}\tan{\left(\varphi\right)}+B^{2}\tan{\left(\phi\right)}+B^{2}\tan{\left(\varphi\right)}}{B\sec^{2}{\left(\phi\right)}}\\ &=\frac{\left(B^{2}-1\right)\tan{\left(\phi\right)}+\left[\tan^{2}{\left(\phi\right)}+B^{2}\right]\tan{\left(\varphi\right)}}{B\sec^{2}{\left(\phi\right)}}\\ &=\frac{B^{2}-1}{2B}\sin{\left(2\phi\right)}+\frac{B^{2}\cos^{2}{\left(\phi\right)}+\sin^{2}{\left(\phi\right)}}{B}\tan{\left(\varphi\right)}.\\ \end{align}$$

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Given fixed but arbitrary $\left(a,b\right)\in\mathbb{R}^{2}$ such that $0<a\land\sqrt{1+a^{2}}<b$, consider the problem of finding all $\left(B,\phi\right)\in\mathbb{R}_{\neq0}\times\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ satisfying the system of equations

$$\frac{B^{2}-1}{2B}\sin{\left(2\phi\right)}=a,$$

$$\frac{B^{2}\cos^{2}{\left(\phi\right)}+\sin^{2}{\left(\phi\right)}}{B}=b.$$

From the second equation, $0<b\land0<B$. Setting $\beta:=2\arctan{\left(B\right)}$, we have $0<\beta<\pi\land B=\tan{\left(\frac{\beta}{2}\right)}$, and the system of equations can be rewritten as

$$\cos{\left(\beta\right)}\sin{\left(2\phi\right)}=-a\sin{\left(\beta\right)},$$

$$\cos{\left(\beta\right)}\cos{\left(2\phi\right)}=1-b\sin{\left(\beta\right)}.$$

Squaring both equations and adding them together,

$$\cos^{2}{\left(\beta\right)}=\left[-a\sin{\left(\beta\right)}\right]^{2}+\left[1-b\sin{\left(\beta\right)}\right]^{2},$$

$$\implies1-\sin^{2}{\left(\beta\right)}=a^{2}\sin^{2}{\left(\beta\right)}+\left[1-2b\sin{\left(\beta\right)}+b^{2}\sin^{2}{\left(\beta\right)}\right],$$

$$\implies2b\sin{\left(\beta\right)}=\left(1+a^{2}+b^{2}\right)\sin^{2}{\left(\beta\right)},$$

$$\implies\sin{\left(\beta\right)}=\frac{2b}{1+a^{2}+b^{2}}.$$

Then, dividing the first equation by the second we have

$$\tan{\left(2\phi\right)}=\frac{a\sin{\left(\beta\right)}}{b\sin{\left(\beta\right)}-1}=\frac{2ab}{b^{2}-a^{2}-1},$$

$$\implies0<\phi<\frac{\pi}{4}\land1<B\land\frac{\pi}{2}<\beta<\pi.$$

The unique solution to our system of equations is found to be

$$B=\frac{\left[\sqrt{a^{2}+\left(b+1\right)^{2}}+\sqrt{a^{2}+\left(b-1\right)^{2}}\right]^{2}}{4b}\land\phi=\frac12\arctan{\left(\frac{2ab}{b^{2}-a^{2}-1}\right)}.$$

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Suppose $\left(a,b,\psi,\omega\right)\in\mathbb{R}\times\mathbb{R}\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\times\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, and assume $0<a\land\sqrt{1+a^{2}}<b\land-\frac{\pi}{4}\le\psi<\omega\le\frac{\pi}{4}$. Set

$$B:=\frac{\left[\sqrt{a^{2}+\left(b+1\right)^{2}}+\sqrt{a^{2}+\left(b-1\right)^{2}}\right]^{2}}{4b}\land\phi:=\frac12\arctan{\left(\frac{2ab}{b^{2}-a^{2}-1}\right)},$$

and note that $1<B\land0<\phi<\frac{\pi}{4}$. We then have

$$\forall\varphi\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]:\arctan{\left(a+b\tan{\left(\varphi\right)}\right)}=-\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\arctan{\left(B\tan{\left(\phi+\varphi\right)}\right)}.$$

Finally,

$$\begin{align} \mathcal{K}{\left(a,b,\psi,\omega\right)} &=\int_{\psi}^{\omega}\mathrm{d}\varphi\,\arctan{\left(a+b\tan{\left(\varphi\right)}\right)}\\ &=-\int_{\psi}^{\omega}\mathrm{d}\varphi\,\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\int_{\psi}^{\omega}\mathrm{d}\varphi\,\arctan{\left(B\tan{\left(\phi+\varphi\right)}\right)}\\ &=-\left(\omega-\psi\right)\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\int_{\psi}^{\omega}\mathrm{d}\varphi\,\arctan{\left(B\tan{\left(\phi+\varphi\right)}\right)}\\ &=-\left(\omega-\psi\right)\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\int_{\phi+\psi}^{\phi+\omega}\mathrm{d}\varphi\,\arctan{\left(B\tan{\left(\varphi\right)}\right)};~~~\small{\left[\varphi\mapsto\varphi-\phi\right]}\\ &=-\left(\omega-\psi\right)\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}+\int_{0}^{\phi+\omega}\mathrm{d}\varphi\,\arctan{\left(B\tan{\left(\varphi\right)}\right)}\\ &~~~~~-\int_{0}^{\phi+\psi}\mathrm{d}\varphi\,\arctan{\left(B\tan{\left(\varphi\right)}\right)}\\ &=-\left(\omega-\psi\right)\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}\\ &~~~~~+\frac12\left(\phi+\omega\right)^{2}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-B}{1+B}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\phi-2\omega\right)}\\ &~~~~~-\frac12\left(\phi+\psi\right)^{2}-\frac12\operatorname{Li}_{2}{\left(-\frac{1-B}{1+B}\right)}+\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\phi-2\psi\right)}\\ &=\frac12\left(\phi+\omega\right)^{2}-\frac12\left(\phi+\psi\right)^{2}-\left(\omega-\psi\right)\arctan{\left(\frac{\tan{\left(\phi\right)}}{B}\right)}\\ &~~~~~-\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\phi-2\omega\right)}+\frac12\operatorname{Li}_{2}{\left(\frac{1-B}{1+B},\pi-2\phi-2\psi\right)}.\blacksquare\\ \end{align}$$

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