# Closed form of :$\int_{-\infty}^{\infty}\arctan\left(e^{-x^2 \text{erf}(x)}\right)\,\arctan\left(e^{x^2\text{erf(x)}}\right)\,dx$

That is one of interesting integral that i have accrossed in my text book when i have tried to understand some thing related to distribution theory in the statistics context , The following integral really make me tired to get its closed form however i find some integral connecting to that they have closed form .

$$\begin{eqnarray*} \int_{-\infty}^{\infty}\arctan\left(e^{-x^2 \text{erf}(x)}\right)\,\arctan\left(e^{x^2\text{erf(x)}}\right)\,dx \sim \frac{5}{\pi} \end{eqnarray*}$$

Now if we use the integration by part as the first step we should accross to get the following integral where i have got it's series representation and it is defined as follow by :

$$$$\int\limits_{0}^{x}e^{-\xi ^{2}\text{erf}(\xi )d\xi }=\sum\limits_{n=0}^{\infty }\lim_{\varepsilon ->0}\left( \sum\limits _{\substack{ k_{1}+2k_{2}+\cdots +nk_{n}=n \\ k_{1}\geq 0,k_{2}\geq 0,...,k_{n}\geq 0}}\prod\limits_{j=1}^{n}\frac{^{A_{j,\varepsilon }^{k_{j}}}% }{k_{j}!}\right) \frac{x^{n+1}}{n+1}$$$$ where: $$\begin{eqnarray*} A_{j,\epsilon } &=&\frac{2(-1)^{(j-1)/2}}{(j-2)(\frac{1}{2}(j-3))!\sqrt{\pi }% }\text{ if \ }j\geq 3\text{ and }j\text{ an odd integer;} \\ A_{j,\epsilon } &=&\varepsilon \text{ otherwise \ }(0<\varepsilon <1)\text{. } \end{eqnarray*}$$ Really that represnation can't give me the result because it is hard to conclude arctan of that complicated series , All my attempt can't give me the closed form , Only i know that integral could be close to $$\frac{5}{\pi}$$ close to the result shown by Wolfram alphasince i know that erf is the function which deals with $$\pi$$ incrementation , Now my question is :

Question: Is it possible to get its closed form value ? and if yes could we get also its series representation ?

• Now what textbook would that be? – omegadot Mar 31 at 1:48