The integral $\int_{0}^{\pi/2} \sin 2\theta ~ \mbox{erf}(\sin \theta)~ \mbox{erf}(\cos \theta)~d\theta=e^{-1}$ In a work it was required to find an integral akin to  $$I=\int_{0}^{\pi/2} \sin 2 \theta~\tanh(\sin\theta) \tanh(\cos \theta)~d\theta.$$ Since $\tanh x$ and $\operatorname{erf}(x)$ are similar functions so it was a pleasure to see that its variant namely  $$J=\int_{0}^{\pi/2} \sin 2\theta ~ \operatorname{erf}(\sin \theta)~ \operatorname{erf}(\cos \theta)~d\theta=e^{-1}$$
could be solved by Mathematica.
So the question is: Can one do it ($J$) by hand?
 A: The antiderivative can be found by hand.  Integrate by parts
$$\newcommand{\erf}{\operatorname{erf}}
\mathrm{d}u=\sin 2\theta\,\mathrm{d}\theta,\quad v=\erf(\cos \theta)\erf(\sin \theta)
$$
gives
$$
-\frac12\cos 2\theta\erf(\sin \theta)\erf(\cos \theta)+
\frac1{\sqrt\pi}\int\cos(2\theta)\left[e^{-\sin^2 \theta}\cos \theta\erf(\cos \theta)-e^{-\cos^2 \theta}\sin \theta\erf(\sin \theta)\right]\,\mathrm{d}\theta
$$
Hence we are reduced to computing
$$
\int\cos(2\theta)\left[e^{-\sin^2 \theta}\cos \theta\erf(\cos \theta)-e^{-\cos^2 \theta}\sin \theta\erf(\sin \theta)\right]\,\mathrm{d}\theta\\
=e^{-1}\int\cos(2\theta)\left[e^{\cos^2 \theta}\cos \theta\erf(\cos \theta)-e^{\sin^2 \theta}\sin \theta\erf(\sin \theta)\right]\,\mathrm{d}\theta
$$
But note that
\begin{align*}
&\frac{\mathrm{d}}{\mathrm{d}\theta}\left[e^{\cos^2 \theta}\sin \theta\erf(\cos \theta)\right]\\
&=(-2\sin \theta\cos \theta)e^{\cos^2 \theta}\sin \theta\erf(\cos \theta)
+e^{\cos^2 \theta}\cos \theta\erf(\cos \theta)
+e^{\cos^2 \theta}\sin \theta\frac2{\sqrt\pi}e^{-\cos^2 \theta}(-\sin \theta)\\
&=(-2\sin^2 \theta+1)e^{\cos^2 \theta}\cos \theta\erf(\cos \theta)
-\frac{2\sin^2 \theta}{\sqrt\pi}\\
&=\cos (2 \theta)e^{\cos^2 \theta}\cos \theta\erf(\cos \theta)
-\frac{2\sin^2 \theta}{\sqrt\pi}
\end{align*}
and the similar equation with $\cos \theta$ and $\sin \theta$ interchanged.
So putting these two together,
\begin{align*}
&\int\cos(2\theta)\left[e^{\cos^2 \theta}\cos \theta\erf(\cos \theta)-e^{\sin^2 \theta}\sin \theta\erf(\sin \theta)\right]\,\mathrm{d}\theta\\
&=\left( e^{\cos^2 \theta} \erf(\cos \theta) \sin \theta - e^{\sin^2 \theta} \erf(\sin \theta) \cos \theta\right)+\int\frac2{\sqrt\pi}(\sin^2 \theta+\cos^2 \theta)\,\mathrm{d}\theta
\end{align*}
and of course you know how to do the final integral.
The final upshot is
\begin{align*}
&\int\sin(2\theta)\erf(\sin\theta)\erf(\cos\theta)\,\mathrm{d}\theta\\
&=\frac2{e\pi}\theta-\frac12\cos 2\theta\erf(\sin \theta)\erf(\cos \theta)\\
&\quad+\frac1{e\sqrt\pi}
\left( e^{\cos^2 \theta} \erf(\cos \theta) \sin \theta - e^{\sin^2 \theta} \erf(\sin \theta) \cos \theta\right)
\end{align*}

For $I$, this approach doesn't work, and I don't think there is a closd form solution to the definite integral.  However, we can express the MacLaurin series
$$
\tanh z=\sum_{k\geq 0} 2\frac{(-1)^k}{\pi^{2k+2}}\left(1-\frac1{4^{k+1}}\right)\zeta(2k+2)z^{2k+1}
$$
so
$$
I=4\sum_{k,k'\geq 0}(1-4^{-k-1})(1-4^{-k'-1})\frac{(-1)^{k+k'}\zeta(2k+2)\zeta(2k'+2)\Gamma(k+\frac12)\Gamma(k'+\frac12)}{\Gamma(k+k'+1)\pi^{2(k+k')+4}}
$$
which doesn't really simplify the task.
A: $$J=\int_{0}^{\pi/2} \sin 2\theta  \mbox{erf}(\sin \theta) \mbox{erf}(\cos \theta). $$
I get clue from the fact that though $\tanh x$ and erf$(x)$ are similar, but the latter one enjoys beautiful expansions!
Expanding $$\mbox{erf}(x)= e^{-x^2} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{\Gamma(n+3/2)}.$$ We re-write
$$J=2e^{-1}\sum_{m=0}^{\infty} \sum _{n=0}^{\infty} \int_{0}^{\pi/2} \frac{ \sin^{2m+2} \theta ~\ cos^{2n+2} \theta} {\Gamma(m+3/2)~\Gamma(n+3/2)} ~d \theta.$$
By Beta-integral, we get
$$J=e^{-1}\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{\Gamma(m+n+2)}=e^{-1}\sum_{k=1}^{\infty} \sum_{m=0}^{k-1} \frac{1}{(m+1)!}=e^{-1}\sum_{k=1}^{\infty} \frac{k}{(k+1)!}.$$
$$\Rightarrow e^{-1} \sum_{k=0} \left( \frac{1}{k!}-\frac{1}{(k+1)!} \right)=e^{-1}.$$
