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

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    $\begingroup$ Wolfram Alpha doesn't recognize "Phi[x]" as a cumulative normal distribution. It does, however, recognized "Erf[x]" and you can show that $2 \Phi(x \sqrt{2}) = \text{Erf}(x) + 1/2.$ wolframalpha.com/input/… $\endgroup$ – Bridgeburners Feb 14 '18 at 22:00
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    $\begingroup$ Actually, I guess it does recognize "Phi[x]" when you input it on its own. But for some reason it registers it as some arbitrary function when you input it into the integral. But it recognizes "Erf[x]". I guess that's because it's a base function in Mathematica. In Wolfram Alpha, when all else fails, it's safe to enter raw Mathematica code. $\endgroup$ – Bridgeburners Feb 14 '18 at 22:08
  • $\begingroup$ Would you be able to comment on how evaluating this integral would reveal something about the "relationship between the error function and the CDF of the normal distribution"? What more is there to determine beyond the equation provided in the first comment by @Bridgeburners? $\endgroup$ – whuber Feb 15 '18 at 23:23
  • $\begingroup$ we have :$\text{Erf}(x) = 2\Phi(x\sqrt{2}) - 1.$ , then Multiplying both side by $-x^2 $ we get this : $-x^2\text{Erf}(x) = -x^2( 2\Phi(x\sqrt{2}) - 1.)$ which it is :$-x^2\text{Erf}(x) =x^2(1 -2(\Phi(x\sqrt{2}) )$ ,Now just to raise exp for both side and integrating both side over positive real line $(0,\infty)$ we w'd get the identity $\endgroup$ – zeraoulia rafik Feb 15 '18 at 23:29
  • $\begingroup$ Small mistake on my part. I did the integral from $-\infty$ to $0$ incorrectly and got $1/2$ when I should have gotten $1.$ So I should have said $2 \Phi(x \sqrt{2}) = \text{Erf}(x) + 1.$ $\endgroup$ – Bridgeburners Feb 15 '18 at 23:38
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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
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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 :$0.97210699\cdots$

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    $\begingroup$ Nice to see that answer from Wolfram Alpha and R agree ... $\endgroup$ – kjetil b halvorsen Feb 16 '18 at 15:13

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