In a generalization of this problem: Integral involving product of arctangent and Gaussian, I am trying to calculate the integral
$$ I(a,b) = \int_{\mathbb{R}^2} \arctan^2 {\left( \frac{y+b}{x+a} \right)} e^{- (x^2 + y^2)} d x d y , $$
for $a,b \in \mathbb{R}$ in terms of commonly used special functions. The substitution that worked in the aforementioned question, i.e.
$$ \frac{1}{(x+a) \sqrt{\pi}} \arctan{ \left( \frac{y+b}{x+a} \right) } = \int_0^{\infty} \mathrm{erf} ((y+b)s) e^{-(x+a)^2 s^2} d s , $$
where $\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} d t$ is the standard error function, forces us to consider the integral
$$ \int_{\mathbb{R}} \mathrm{erf} ((y+b)t) \mathrm{erf} ((y+b)s) e^{- y^2} d y . $$
In the case of $b = 0$, we may use the result
$$ \int_0^{\infty} \mathrm{erf}{(ax)} \mathrm{erf}{(bx)} e^{-c^2 x^2} d x = \frac{1}{c \sqrt{\pi}} \arctan{\left( \frac{ab}{c \sqrt{a^2 + b^2 + c^2}} \right)} $$
found in, e.g., formula (18) of page 158 of https://nvlpubs.nist.gov/nistpubs/jres/75B/jresv75Bn3-4p149_A1b.pdf. Making this substitution, we find
$$ I(a,0) = \pi \int_0^{\pi/2} \int_0^{\infty} r \arctan{ \left( \frac{r \cos{\theta}\sin{\theta}}{\sqrt{1+r^2}} \right) } \exp{ \left\lbrace – \frac{a^2 r^2}{1 + r^2} \right\rbrace} \frac{(1 + 2 a^2 + r^2)}{(1+r^2)^{5/2}} d \theta d r . $$
I would be very happy just to demonstrate a closed-form for $I(a,0)$ above. However, now I am stuck trying to compute
\begin{align}
\int_0^{\pi / 2} \arctan{ \left( \frac{r \cos{\theta}\sin{\theta}}{\sqrt{1+r^2}} \right) } d \theta . \hspace{2cm} (*)
\end{align}
Would anyone on this platform know how to compute ($*$)? I have tried Mathematica, but it returns some nasty expression. Maybe I am going about this wrong by trying to compute the $\theta$ integral first? Any ideas about ($*$) or the original integral $I(a,b)$ are appreciated. I asked about a related integral here: Integral involving sin, cosine, exponential, and error functions.. The integral in this question is found by starting from polar coordinates with $I(a,b)$. Unfortunately, I have not been able to make progress on this integral either.
Update: Observe that
$$ \frac{d}{dr} \frac{e^{- a^2 r^2 / (1+r^2)}}{\sqrt{1+r^2}} = – r \exp{ \left\lbrace – \frac{a^2 r^2}{1 + r^2} \right\rbrace} \frac{(1 + 2 a^2 + r^2)}{(1+r^2)^{5/2}} . $$
We may therefore integrate-by-parts to find
\begin{align*}
I(a,0) & = – \pi \int_0^{\pi/2} \int_0^{\infty} \arctan{ \left( \frac{r \cos{\theta}\sin{\theta}}{\sqrt{1+r^2}} \right) } \frac{d}{dr} \frac{e^{- a^2 r^2 / (1+r^2)}}{\sqrt{1+r^2}} dr d \theta \\
& = \pi \int_0^{\pi/2} \int_0^{\infty} \frac{\cos{\theta} \sin{\theta}}{(1+r^2) (1 + r^2 ( 1 + \cos^2{\theta} \sin^2{\theta} )) } e^{- a^2 r^2 / (1+r^2)} dr d \theta .
\end{align*}
Hopefully this observation simplifies the problem some. Apparently, Mathematica believes
$$ \int_0^{\pi/2} \frac{\cos{\theta} \sin{\theta}}{1 + r^2 ( 1 + \cos^2{\theta} \sin^2{\theta} ) } d \theta = \frac{2 \mathrm{arctanh}{ \frac{r}{\sqrt{5 r^2 + 4}} }}{r \sqrt{ 5 r^2 + 4 } } , $$
where $\mathrm{arctanh}$ is the inverse hyperbolic tangent. I haven't been able to derive this myself though.
Update 2: I was able to make a little more progress on the special case $I(a,0)$. Starting from the previous identity involving the hyperbolic $\mathrm{arctanh}$, our integral of interest becomes
\begin{align*}
I(a,0) = 2 \pi \int_0^{\infty} \frac{2 \mathrm{arctanh}{ \frac{r}{\sqrt{5 r^2 + 4}} }}{r (1 + r^2) \sqrt{ 5 r^2 + 4 } } e^{- a^2 r^2 / (1+r^2)} d r .
\end{align*}
Now, observe that
\begin{align*}
\frac{d }{d r} \mathrm{arctanh}^2{ \frac{r}{\sqrt{5 r^2 + 4}} } = 2 \frac{\mathrm{arctanh}{ \frac{r}{\sqrt{5 r^2 + 4}} }}{(1+r^2) \sqrt{5 r^2 + 4}} .
\end{align*}
Then, via integration-by-parts,
\begin{align*}
I(a,0) & = 2 \pi \int_0^{\infty} \frac{1}{r} e^{- a^2 r^2 / (1+r^2)} \frac{d }{d r} \mathrm{arctanh}^2{ \frac{r}{\sqrt{5 r^2 + 4}} } d r \\
& = 2 \pi \int_0^{\infty} \left( \frac{1}{r^2} + \frac{2a^2}{(1+r^2)^2} \right) e^{- a^2 r^2 / (1+r^2)} \mathrm{arctanh}^2{ \frac{r}{\sqrt{5 r^2 + 4}} } d r .
\end{align*}
Now we look to making the Euler substitution to simplify the integral. Let $\sqrt{ 5 r^2 + 4 } = \sqrt{5} r + t$. Then,
\begin{align*}
r = \frac{4 – t^2}{2 \sqrt{5} t} = \frac{\sqrt{5}}{10} \left( \frac{4}{t} – t \right).
\end{align*}
Consequently,
\begin{align*}
d r = – \frac{\sqrt{5}}{10} \left( \frac{4}{t^2} + 1 \right) d t .
\end{align*}
Plugging this into our integral and simplifying reveals
\begin{align}
I(a,0) = 4 \pi \sqrt{5} \int_{- \infty}^{\infty} (4 + t^2) \left( \frac{1}{(t^2 – 4)^2} + \frac{40 a^2 t^2}{(t^4 + 12 t^2 + 16)^2} \right) e^{- \frac{a^2 (t^2 – 4)^2}{16+12t+t^2}} \mathrm{arctanh}^2{ \left( \frac{t^2 – 4}{ \sqrt{5} (t^2 + 4)} \right) } d t .
\end{align}
It is starting to seem as though some "closed-form" in terms of $a$ is possible. Due to the appearance of $\mathrm{arctanh}^2$ in the previous integral, I am inclined to conjecture that $\mathrm{arctanh}$ is part of the final answer. This would be a nice "duality" with this integral and the final answer to the related one in the question linked at the very top.
Does anyone have any suggestions on how I could proceed?
Update 3: I found a mistake in my original version of ($\ast$). Please see the posted answer of mine for an update.
Best Answer
Using $k=\frac r {2\sqrt{1+r^2}}$ $$I=\int_0^{\pi / 2} \arctan{ \left( \frac{r \cos(\theta)\sin(\theta)}{\sqrt{1+r^2}} \right) }\, d \theta =\int_0^{\pi / 2} \arctan{ \left(k \sin(2\theta)\right) }\, d \theta$$ $$I=\text{Li}_2\left(k-\sqrt{k^2+1}\right)-\text{Li}_2\left(-k+\sqrt{k^2+1}\right)-2 \sinh ^{-1}(k) \coth ^{-1}\left(\sqrt{k^2+1}+k\right)+\frac{\pi ^2}{4}$$ The result does not look so bad.
As a function of $k$ $(0 \leq k \leq \frac 12)$, it is very close to a straight line. Expanded as a series $$I=k-\frac{2 }{9}k^3+\frac{8 }{75}k^5-\frac{16}{245} k^7+\frac{128 }{2835} k^9+O\left(k^{11}\right)$$