How can I show that $\int_0^{\frac{\pi}{2}}\sin\left(\frac{x}{2}\right) \text{arctanh}\left(\sin(2x)\right) dx$ has this value?

Question

Question: How can I show that \begin{align} & \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \text{arctanh}\left(\sin(2x)\right)\,dx \\[2mm] = & \ {\small\log\left(\left(2\sqrt{2-\sqrt{2}}+2\sqrt{2}-1\right)^{\sqrt{2+\sqrt{2}}} \left(1+2\sqrt{2}-2\sqrt{2+\sqrt{2}}\right)^{\sqrt{2-\sqrt{2}}}\right)} \end{align}

My attempt

I will rewrite $\text{arctanh}(\sin(2x))$ in terms of logarithms.

We know that $$ \text{arctanh}(y) = \frac{1}{2} \ln\left(\frac{1+y}{1-y}\right) $$

Here $y=\sin(2x)$ $$ \text{arctanh}(\sin(2x)) = \frac{1}{2} \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) $$

Using trigonometric identities, we can write: $$ 1 + \sin(2x) = 1 + 2 \sin(x) \cos(x) $$ $$ 1 - \sin(2x) = 1 - 2 \sin(x) \cos(x) $$

or
$\sin(2x) = 2 \sin(x) \cos(x)$

Substitute $\text{arctanh}(\sin(2x))$: $$ \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \text{arctanh}(\sin(2x)) \, dx \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \cdot \frac{1}{2} \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) \ dx $$

Simplifying the logarithm expression:

Using the given trigonometric identities: $$ 1 + \sin(2x) = \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2 $$ $$ 1 - \sin(2x) = \left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2 $$

Thus, $$ \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) = \ln\left(\frac{\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2}{\left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2}\right) $$

so our integral becomes

$$ \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \ln\left(\frac{\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2}{\left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2}\right)\, dx $$

Answer 1

Here's another way to evaluate the integral, overlapping slightly with Julio's suggestion:

$$\begin{align*} & \int_0^\tfrac\pi2 \sin\frac x2 \, \operatorname{artanh}(\sin(2x)) \, dx \\ &= \left\{\int_0^\tfrac\pi4 + \int_\tfrac\pi4^\tfrac\pi2\right\} \sin\frac x2 \, \log\left\lvert \sec(2x) + \tan(2x)\right\rvert \, dx \\ &= \int_0^\tfrac\pi4 \sin\frac x2 \, \log(\sec(2x)+\tan(2x)) \, dx \\ &\qquad + \int_\tfrac\pi4^\tfrac\pi2 \sin\frac x2 \, \log(-\sec(2x)-\tan(2x)) \, dx \\ &= \int_0^\tfrac\pi4 \sin\frac x2 \, \log \frac{1+\tan x}{1-\tan x} \, dx \\ &\qquad + \int_0^\tfrac\pi4 \sin\frac{x+\frac\pi4}2 \, \log(\cot x) \, dx & x\to x+\frac\pi4 \\ &= \int_0^\tfrac\pi4 \left(\sin\frac{x+\frac\pi4}2 - \sin\frac{x-\frac\pi4}2\right) \log(\cot x) \, dx & x\to\frac\pi4-x \\ &= -2\sin\frac\pi8 \int_0^\tfrac\pi4 \cos\frac x2 \, \log(\tan x) \, dx \\ &= 8\sin\frac\pi8 \int_0^\tfrac\pi4 \frac{\sin\frac x2}{\sin(2x)} \, dx & \rm IBP \\ &= 16\sin\frac\pi8 \int_0^\tfrac\pi8 \frac{\sin x}{\sin(4x)} \, dx & x\to2x \\ &= 4\sin\frac\pi8 \int_0^\tfrac\pi8 \left(\frac{2\cos x}{1-2\sin^2x} - \sec x\right) \, dx \\ &= 4\sin\frac\pi8 \left\{\sqrt2 \int_0^{\arcsin\left(\sqrt2\sin\tfrac\pi8\right)} - \int_0^\tfrac\pi8\right\} \sec x \, dx & \sin x\to\frac1{\sqrt2}\sin x \\ &= \frac{4a}{\sqrt{1+a^2}} \left[(1+a) \log\left(\sqrt a + \sqrt{1+a}\right) - \log\left(\sqrt{1+a^2}+a\right)\right] \\ \end{align*}$$

where $a = \tan\dfrac\pi8 = \sqrt2-1$.

Showing it's equivalent to the given closed form is another matter, but this result does agree with the definite integral. (WA has problems with the singularity at $\pi/4$)

Answer 2

There are probably easier ways to find the definite integral, for instance, using coangle formulas or symmetry, as already mentioned by others. Nonetheless, it is possible to find an antiderivative for the integrand.

Just for simplicity, let $u=\dfrac{x}{2}$.

Then, $$\int\sin\left(\dfrac{x}{2}\right)\operatorname{arctanh}(\sin(2x))\,dx=2\int\sin\left(u\right)\operatorname{arctanh}(\sin(4u))\,du$$

Integration by parts yields $$\int\sin\left(u\right)\operatorname{arctanh}(\sin(4u))\,du=-\cos(u)\operatorname{arctanh}(\sin(4u))+\int\dfrac{4\cos(u)\cos(4u)}{1-\sin^2(4u)}\,du=$$$$-\cos(u)\operatorname{arctanh}(\sin(4u))+\int \dfrac{4\cos(u)}{\cos(4u)}\,du$$

Now, using that $\cos(4u)=8(\sin^4(u)-\sin^2(u))+1$ and substituting $v=\sin(u)$, the last integral is the same as $$\int\dfrac{1}{8(v^4-v^2)+1}\,dv$$

This can be decomposed on partial fractions, and we get immediate logarithm integrals. Reverse engineering all these steps, we get an antiderivative for $\sin\left(\dfrac{x}{2}\right)\operatorname{arctanh}(\sin(2x))$. The definite integral can then be computed by Barrow's Rule.

As a small way of seeing this process is related to the value you are getting, notice that $\dfrac{\sqrt{2-\sqrt{2}}}{2}$ is a root of $8(v^4-v^2)+1$.

Answer 3

Assume that $0 < a< 2$ and $|r|<1$.

Let $$I(a,r) = \int_{0}^{\pi/2} \sin(ax) \operatorname{artanh} \left(\frac{2r \sin 2x}{1+r^{2}} \right ) \, \mathrm dx. $$

For $|r| < 1$ and $\theta \in \mathbb{R}$, we have the Fourier series $$2 \sum_{n=0}^{\infty} \frac{r^{2n+1} \sin \left((2n+1) \theta\right)}{2n+1} = \arctan \left(\frac{2r \sin \theta}{1-r^{2}} \right).$$

Therefore, since $\arctan(ix) = i \operatorname{artanh}(x)$, $$ \begin{align} I(a,r) &=-2 \int_{0}^{\pi/2} \sin(ax) \sum_{n=0}^{\infty} \frac{(-1)^{n} r^{2n+1} \sin \left((2n+1)2x \right)}{2n+1} \, \mathrm dx \\ &= -2 \sum_{n=0}^{\infty} \frac{(-1)^{n}r^{2n+1}}{2n+1} \int_{0}^{\pi/2} \sin(ax) \sin \left((2n+1)2x \right) \, \mathrm dx \\ &= \frac{2}{a} \, \sin \left(\frac{\pi a}{2} \right) \sum_{n=0}^{\infty} (-1)^{n} r^{2n+1}\left(\frac{1}{2(2n+1)-a}- \frac{1}{2(2n+1)+a} \right) \\ &= \frac{1}{2a} \, \sin \left(\frac{\pi a}{2} \right)\sum_{n=0}^{\infty} (-1)^{n} r^{2n+1} \left(\frac{1}{n+ \frac{2-a}{4}} - \frac{1}{n+ \frac{2+a}{4}} \right) \\ &= \frac{r}{2a} \, \sin \left(\frac{\pi a}{2} \right)\left(\Phi \left(-r^{2}, 1, \frac{2-a}{4} \right) - \Phi \left(-r^{2}, 1, \frac{2+a}{4} \right)\right), \end{align}$$

where $\Phi(z,s,a)$ is the Lerch transcendent.

Letting $r \to 1^{-}$, we get $$ \begin{align} I(a,1) &= \int_{0}^{\pi/2} \sin(ax) \operatorname{artanh} \left(\sin 2x \right) \, \mathrm dx \\ &= \frac{1}{2a} \, \sin \left(\frac{\pi a}{2} \right) \left(\Phi \left(-1,1,\frac{2-a}{4} \right) - \Phi \left(-1,1,\frac{2+a}{4} \right)\right) \\ &\overset{\clubsuit}{=} \frac{1}{4a} \, \sin \left(\frac{\pi a}{2} \right) \left(\psi \left(\frac{6-a}{8} \right) - \psi \left(\frac{2-a}{8} \right)- \psi \left(\frac{6+a}{8} \right) + \psi \left(\frac{2+a}{8} \right) \right) \\&= \small \frac{1}{4a} \, \sin \left(\frac{\pi a}{2} \right) \left(2 \psi \left( \frac{2+a}{2}\right) - 2 \psi \left(\frac{2-a}{8} \right)+ \pi \cot \left(\pi \, \frac{2+a}{8} \right) - \pi \cot \left(\pi \, \frac{2-a}{8} \right)\right) . \end{align}$$

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$\clubsuit$ https://mathworld.wolfram.com/DigammaFunction.html (5)

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And using Gauss' digamma theorem, we get $$ \begin{align} I \left(\tfrac{1}{2}, 1 \right) &= \int_{0}^{\pi/2} \sin \left(\frac{x}{2} \right) \operatorname{artanh}(\sin 2x) \, \mathrm dx \\ &= \sqrt{2} \, \sum_{n=1}^{7} \left( \cos \left(\frac{5 \pi n}{8} \right) \ln \sin \left(\frac{\pi n}{16} \right) - \cos \left(\frac{3 \pi n}{8} \right) \ln \sin \left(\frac{\pi n}{16} \right) \right) \\ &= 2 \sqrt{2} \left(\cos \left(\frac{\pi}{8} \right) \ln \tan \left(\frac{3 \pi}{16} \right) + \sin \left(\frac{\pi}{8} \right) \ln \cot \left(\frac{\pi}{16} \right) \right) \\&= 0.694334 \ldots \end{align}$$

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The formula holds for all $a\in \mathbb{R}$ if the right side of the equation is treated as a limit.