Prove that:
$$\int_{0}^{1}\dfrac{x}{1+x^4}\,\arctan(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\,\approx\,0.211322\dots$$
Also prove that:
$$\int_{0}^{1}\dfrac{x^3}{1+x^4}\,\text{artanh}(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\,\approx\, 0.211322\dots\,=\,\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x) \,\mathrm{d}x$$
Notice that $\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\Longrightarrow$ $\frac{1}{2}\,\operatorname{Re}(\arctan^2(\sqrt{\mathrm{i}}))$
I used a very long and probably wrong method, following the technique used in one of my previous posts: The definite Integral $I=\int_{0}^{1}\frac{x\cdot \operatorname{artanh} x}{1+x^2} dx$
Basically one can recreate this integral by replacing $x=\sqrt{\mathrm{i}}$ in the $\arctan^2(x)$ expansion and equating the real part to the corresponding infinite G.P generated paralelly… (I’m sorry for being so vague, I’ve been off my medications for a week)
I would greatly appreciate any alternate derivations 🙂
My derivation:
Consider: $$\arctan(\sqrt{i})=\sqrt{i}-\dfrac{\sqrt{i}^3}{3}+\dfrac{\sqrt{i}^5}{5}-\dfrac{\sqrt{i}^7}{7}+\cdots=\sqrt{i}\left(\left(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{9}-\cdots\right)-\left(\dfrac{i}{3}-\dfrac{i}{7}+\dfrac{i}{11}-\cdots\right)\right)$$
WKT: $\arctan(\sqrt{i})=\dfrac{\pi}{4}+i\dfrac{\ln{(\sqrt{2}+1})}{2}$ and $\sqrt{i}=\dfrac{i+1}{\sqrt{2}}$
From these identities it is easy to extrapolate the identities:
$$\dfrac{\pi}{4\sqrt{2}}+\dfrac{\ln(\sqrt{2}+1)}{2\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{9}-\dfrac{1}{13}+\cdots$$ And:$$\dfrac{\pi}{4\sqrt{2}}-\dfrac{\ln(\sqrt{2}+1)}{2\sqrt{2}}=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{11}-\dfrac{1}{15}+\cdots$$
Multiplying theses two gives $$\dfrac{\pi^2}{16}-\dfrac{\ln^2{(\sqrt{2}+1)}}{4}=\dfrac{1}{1\cdot3}-\dfrac{1}{1\cdot7}+\cdots-\dfrac{1}{5\cdot3}+\dfrac{1}{5\cdot7}-\cdots$$
This sequence can be alternatively derived from: $$\int_{0}^{1}{\sum_{n=1}^{\infty}(-1)^{n+1}x^{4n+1}\arctan(x) dx}=\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x) dx$$
Therefore,$$\int_{0}^{1}\dfrac{x}{1+x^4}\,\arctan(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)$$
I might have made a few mistakes somewhere… Also I would like to know if these kinds of integrals have a particular name.
Best Answer
Here's a fun proof for the second part of your question. (Your conjecture is off by a factor of $1/2$.)
$$ \bbox[15px,#E6FFF3 ,border:5px groove #FF8B00 ]{\int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx=\frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)} $$
Proof. Let the integral equal $\mathcal{I}$. Then map $x \mapsto \tanh x$ so that
$$ \begin{align} \mathcal{I} &:= \int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx \\ &= \int_{0}^{\infty}\frac{x\tanh\left(x\right)^{3}}{1+\tanh\left(x\right)^{4}}\operatorname{sech}^{2}xdx \\ &= \int_{0}^{\infty}\frac{2xe^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx \\ &= \int_{\mathbb{R}}\frac{xe^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx \\ \end{align} $$
where in the last line, we use the fact that the integrand is an even function.
Let $z \mapsto \displaystyle \frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)}$ and define it by the holomorphic function $\displaystyle f: \mathbb{C} \backslash \left\{n \in \mathbb{Z}: z_1(n), z_2(n), z_3(n)\right\} \to \mathbb{C}$. The simple poles excluded from the domain are
$$ \begin{align} z_{1}(n)&=\frac{i\pi}{2}\left(2n+1\right) \\ z_{2}(n)&=\frac{1}{4}\ln\left(3-2\sqrt{2}\right)+\frac{i\pi}{4}\left(2n+1\right) \\ z_{3}(n)&=\frac{1}{4}\ln\left(3+2\sqrt{2}\right)+\frac{i\pi}{4}\left(2n+1\right) \end{align} $$
which are obtained from finding the zeroes of $\displaystyle \frac{1}{f}$. We also construct a rectangular box contour $C$ with vertices $(-R,0)$, $(R,0)$, $(R,R+i\pi)$, and $(-R,-R+i\pi)$ where $R$ is sufficiently large such that $R \gg \displaystyle \frac{1}{4}\ln(3+2\sqrt{2})$. Below, I provide a visual I made of what $C$ looks like in the phase plot of $f$ with domain coloring and shading.
We write each contribution of the integral over $C$ as
$$ \require{cancel}\oint_C f(z)dz=\cancelto{0}{\int_{-R}^{R}f(x)dx}+\int_{R}^{R+i\pi}f(z)dz+\int_{R+i\pi}^{-R+i\pi}f(z)dz+\int_{-R+i\pi}^{-R}f(z)dz $$
where the integral over $[-R,R]$ vanishes because the integrand is an odd function.
Taking $R\to\infty$ on both sides, we get
$$ \require{cancel} \lim_{R\to\infty}\oint_C f(z)dz = \cancelto{0}{\lim_{R\to\infty}\int_{R}^{R+i\pi}f(z)dz} + \lim_{R\to\infty}\int_{R+i\pi}^{-R+i\pi}f(z)dz + \cancelto{0}{\lim_{R\to\infty}\int_{-R+i\pi}^{-R}f(z)dz}\,. $$
The integral over the vertical line segment on the right of $C$ vanishes due to the Estimation Lemma and the Squeeze Theorem, and the other integral over the left vertical segment also vanishes for the same kind of reason, I think.
To prove the integral over the vertical line segment on the right of the box decays to $0$ as $R \to \infty$, we bound the modulus of $f(z)$ first like
$$ \begin{align} |f(z)| &= \left|\frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)}\right| \\ &\leq \frac{\left|z\right|^{2}\cdot\left|e^{2z}\right|\cdot\left(3\left|e^{z}\right|+3\left|e^{2z}\right|+3\left|e^{3z}\right|+1\right)^{2}}{\left|\left|e^{2z}\right|-1\right|\cdot\left|6\left|e^{4z}\right|-\left|e^{8z}\right|-1\right|}\,. \\ \end{align} $$
Parameterizing $z=R+iy$ for $y \in [0,\pi]$, we get
$$ \begin{align} |f(R+iy)| &\leq \frac{\left|R+iy\right|^{2}\cdot\left|e^{2\left(R+iy\right)}\right|\cdot\left(3\left|e^{\left(R+iy\right)}\right|+3\left|e^{2\left(R+iy\right)}\right|+3\left|e^{3\left(R+iy\right)}\right|+1\right)^{2}}{\left|\left|e^{2\left(R+iy\right)}\right|-1\right|\cdot\left|6\left|e^{4\left(R+iy\right)}\right|-\left|e^{8\left(R+iy\right)}\right|-1\right|} \\ &\leq \frac{e^{2R}\left(R^{2}+2Ry+y^{2}\right)\left(3e^{R}+3e^{2R}+3e^{3R}+1\right)^{2}}{\left(e^{2R}-1\right)\left(e^{8R}-6e^{4R}+1\right)} \\ &:= M(R)\,. \\ \end{align} $$
As $R \to \infty$, we see that $M(R)\to 0$ because the largest expression in the numerator grows slower than that of the denominator. Using the Estimation Lemma, we write
$$ 0 \leq \left|\int_{R}^{R+i\pi} f(z)dz\right| \leq \pi M(R)\,. $$
The $\pi$ comes from the length of the right side of the box contour. Taking $R \to \infty$ and employing the Squeeze Theorem, we get
$$ \lim_{R \to \infty} 0 \leq \lim_{R \to \infty} \left|\int_{R}^{R+i\pi} f(z)dz\right| \leq \pi \lim_{R \to \infty} M(R) $$ $$ \implies \lim_{R \to \infty}\left|\int_{R}^{R+i\pi} f(z)dz\right| = 0\,. $$
Thus,
$$ \bbox[15px,#FFF5FE]{\lim_{R \to \infty}\int_{R}^{R+i\pi} f(z)dz = 0\,.} $$
The other integral should be evaluated in a similar fashion.
For the integral that doesn't vanish, it surprisingly helps us recover the integral we want. Parameterizing $z = x+ i\pi$, we get
$$ \begin{align} \lim_{R\to\infty}\int_{R+i\pi}^{-R+i\pi}f(z)dz &= -\lim_{R\to\infty}\int_{-R}^{R}f(x+i\pi)d(x+i\pi) \\ &= -\lim_{R\to\infty}\int_{-R}^{R}\frac{\left(x+i\pi\right)^{2}e^{2\left(x+i\pi\right)}\left(e^{x+i\pi}-1\right)^{3}\left(e^{x+i\pi}+1\right)^{3}}{\left(e^{2\left(x+i\pi\right)}+1\right)\left(6e^{4\left(x+i\pi\right)}+e^{8\left(x+i\pi\right)}+1\right)}dx \\ \require{cancel} &= -\lim_{R\to\infty}\cancelto{0}{\int_{-R}^{R}f(x)dx} - 2\pi i \mathcal{I} + \pi^2 \lim_{R\to\infty}\cancelto{0}{\int_{-R}^{R}\frac{e^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx}\,. \\ \end{align} $$
The last integral vanishes — thankfully — because its integrand is an odd function. Same for the other one that canceled to $0$.
So far, we have
$$ -\frac{1}{2\pi i} \lim_{R\to\infty} \oint_C f(z)dz = \mathcal{I}\,. $$
But we can proceed further using Cauchy's Residue Theorem and equating the real part on both sides as follows:
$$ -\Re\frac{1}{\cancel{2\pi i}} \cancel{2\pi i} \sum_{n \in \mathcal{P}}\mathop{\mathrm{Res}}_{z=n}f(z) = \Re I = I\,. $$
Here, we will denote $\mathcal{P}$ as the set of poles inside $C$, which is $\left\{z_1(0), z_2(0), z_2(1), z_3(0), z_3(1)\right\}$.
Taking the residue at the simple pole $z_1(0)$ of $f(z)dz$, we have
$$ \begin{align} -\Re \mathop{\mathrm{Res}}_{z=i\pi/2}f(z) &= -\Re\lim_{z \to i\pi/2}\frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\frac{d}{dz}\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)} \\ &= -\Re\lim_{z \to i\pi/2} \frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{2e^{2z}\left(12e^{2z}+18e^{4z}+4e^{6z}+5e^{8z}+1\right)} \\ &= -\Re \frac{\pi^2}{8} \\ &= -\frac{\pi^2}{8}\,. \\ \end{align} $$
You can probably guess that the other residues at the four other elements of $\mathcal{P}$ are a drag to calculate, so I'll omit the details. Basically,
$$ \begin{align} -\Re \mathop{\mathrm{Res}}_{z=z_2(0)}f(z) &= \frac{\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3-2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_2(1)}f(z) &= \frac{9\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3-2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_3(0)}f(z) &= \frac{\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3+2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_3(1)}f(z) &= \frac{9\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3+2\sqrt{2}\right)\,, \\ \end{align} $$
after a lot of brutal computations.
All five residues sum to
$$ \bbox[15px,border:5px groove #FF18F3 ]{-\Re\sum_{n \in \mathcal{P}}\mathop{\mathrm{Res}}_{z=n}f(z) = \frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)\,.} $$
We finally have
$$ \bbox[15px,#FBFFEB ,border:5px groove #00906E ]{\int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx=\frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)} $$
and we're done! $\blacksquare$
Cheers :)