What is the value of $ \int_{0}^{+\infty}e^{x^2(1-2\Phi(x\sqrt{2}))}dx$?

Question

I have accrossed the below integral when i have tried to know more about relationship between error function and CDF of the normal distribution ,I plug this integral in wolfram alpha but no result , but some of my weaker gaven assure that is convergents, then my question here is what is the value of :

$$ \int_{0}^{+\infty}e^{x^2(1-2\Phi(x\sqrt{2}))}dx$$

With :$\Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-z^2/2} \mathrm{d}z.$

Answer 1

If you need a symbolic answer I guess you are out of luck. I would numerical integration. In R:

 f  <-  function(x) exp(x^2*(1-2*pnorm(x*sqrt(2))))
 integrate(f, lower=0, upper=+Inf)
0.972107 with absolute error < 2.2e-06

Answer 2

Hint: just Idea coming up to my mind , just to use the identity gaven above by Bridgeburners which showed the relationship between error function and CDF distribution function , I have got the following identity :

$$ \int_{0}^{+\infty}e^{x^2(1-2\Phi(x\sqrt{2}))}dx=\int_{0}^{+\infty} {(e^{-x²})}^{\text{erf}{(x)}}dx \tag{1}$$.

The RHS of $(1) $is convergent and wolfram alpha says that is :$0.97210699\cdots$