How do I evaluate $\int_{0}^{T}\tanh^{-1}(\sin(x))dx$?

Question

I am trying to solve the following integral,

$$\int_{0}^{T}\tanh^{-1}(\sin(x))dx.$$

I have tried writing $\tanh^{-1}(\sin(x))$ as a series and solving the resultant $\sin^n(x)$ integrals using a recursion formula. I have also tried substituting $\sin(x)=u$. But I am not getting anywhere. Part of my problem is that I am not sure that the integral has a solution and I am not sure how to test this.

Note that I am not a mathematician, nor especially mathematically literate. This came up in a research project and I now have another approach which doesn't use this integral. I just want to know it this integral is soluble, and if so, how? It would also be useful to check my other method using this method.

Answer 1

Wolfram Alpha gives the indefinite integral of your expression as

$$\int \tanh^{-1}(\sin(x)) dx = -i \mathrm{Li}_2(-i e^{i x}) + i \mathrm{Li}_2(i e^{i x}) \\ + x (-\log(1 - i e^{i x}) + \log(1 + i e^{i x}) + \tanh^{-1}(\sin(x))) + c$$

where $\mathrm{Li}_2$ is the polylogarithm function of order 2. This might not be particularly satisfying, but it's also possibly the best that we can do.

Note that the integrand has discontinuities, so you need to be careful if your value of $T$ is greater than $\frac{\pi}{2}$. On that note, however, we can find some specific values of the integral, for example if $T = \frac{\pi}{2}$ then the value is 2C, where $C$ is Catalan's constant.

If you wanted to find the integral by hand, then it would probably involve application of some combination of the following:

• Using the identity $\tanh^{-1}(x) = \frac{1}{2}(\ln(1+u) - \ln(1-u))$

• Using the identity $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

• Integration by parts, particularly via $\int 1 \cdot f(x) dx = x f(x) - \int x f'(x) dx$

Answer 2

There is no closed form in terms of elementary functions, but we can proceed as follows. First, note that the integrand satisfies $$\frac{d}{dx} \operatorname{artanh} \sin x = \sec x$$ ---in fact, $\operatorname{artanh} \sin$ is the anti-Gudermannian or Lambertian function, $\operatorname{gd}^{-1} x = \int_0^x \sec t \,dt$, which appears, e.g., in the construction of the Mercator projection. So, integrating by parts, taking $u = \operatorname{artanh} \sin x$, $dv = dx$, gives $$\int_0^T \operatorname{artanh} \sin x \,dx = T \operatorname{artanh} \sin T - \int_0^T x \sec x \,dx .$$ Per the answers to this question about the latter integral, this expression has value $$\boxed{\int_0^T \operatorname{artanh} \sin x \,dx = T \operatorname{artanh} \sin T + i(\operatorname{Li}_2(ie^{i T})-\operatorname{Li}_2(-ie^{i T}))+ 2 T \arctan (i e^{i T}) + 2 G} ,$$ where $\operatorname{Li}_2$ is the dilogarithm and $G$ is Catalan's constant. The particular value $T = \frac\pi2$ gives the pleasant identity $\int_0^\frac\pi2 \operatorname{artanh} \sin x \,dx = 2 G$.

Answer 3

Via integration by parts, we have

$$ \begin{aligned} \int_0^T \tanh ^{-1}(\sin x) d x = & {[x \tanh (\sin x)]_0^T-\int_0^T \frac{x \cos x}{1-\sin ^2 x} \cdot d x } \\ = & T \tanh ^{-1}(\sin T)-\int_0^T \frac{x}{\cos x} d x \end{aligned} $$

Plugging the result in the post $$ \int\frac{t}{\cos t} \,dt=-2it\arctan(e^{it})+i(\mathrm{Li}_2(-ie^{it})-\mathrm{Li}_2(ie^{it})) $$ into our integral, we have

$$ \begin{aligned} I= & T \tanh ^{-1}(\sin T)-\left[-2 i x \arctan \left(e^{i x}\right) +i\left(\mathrm{Li}_2\left(-i e^{i x}\right)-\mathrm{Li}_2\left(i e^{i x}\right)\right)\right]_0^{\top} \\ = & T \tanh ^{-1}(\sin T)+2 i T \arctan \left(e^{i T}\right)+i\left[\mathrm{Li}_2\left(-i e^{i T}\right)-\mathrm{Li}_2\left(i e^{i T}\right)\right] -i\left[L_{i_2}(-i)-L_{i_2}(i)\right] \\ = & T \tanh ^{-1}(\sin T)+2 i T \arctan \left(e^{i T}\right)+i\left[\mathrm{Li}_2\left(-i e^{i T}\right)-\mathrm{Li}_2\left(i e^{i T}\right)\right] +2 G \end{aligned} $$ where $\mathrm{Li}_2(.)$ is the dilogarithm function and $G$ is the Catalan constant.