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$\newcommand{\erf}{\operatorname{erf}}$ The computation of $\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx$ for large $a$ gives $0.972106...$ by wolfram alpha, but according to JJacquelin comments which claimed that $I(a)=\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx$ is a non-standard special function ,and according to the comments of yuriy which showed that is a just a constant , then it must to ask about closed form of

$$I(a)=\int_{0}^a {(e^{-x²})}^{\operatorname{erf}(x)}dx $$ if it exists with $a$ positive real number , and also to know if this special function behave as the same with Error function or exponontial function since it is a composition of them ? and what about it's application in probability and in differential equation area ? In addition of that what about representation of it's hypergeometric function ?

Edit 01: I have edited the question according to the below comments and to know the closed form of this special function since it is not standard .

Edit 02 I have added some detaill in my question according to the good answer which is gaven by yuriy in order to know more about this new function

Edit: I have added other question which seking for the hypergeometric function of the titled function .

Note (01):$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt.$

Note (02): The motivation of this function is to look for the behavior of integral of some functions where they are represented as a function power it's antiderivative function however the example we take here is :$\int ({f'})^{kf}$ .

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    $\begingroup$ Why do you think a closed form exists? $\endgroup$ – Yuriy S Feb 10 '18 at 9:05
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    $\begingroup$ May I ask how you did arrive to this integral ? $\endgroup$ – Claude Leibovici Feb 10 '18 at 9:29
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    $\begingroup$ Thanks to the special function :$$Z_{\text{era}}(x)=\int_0^x \left(e^{-t^2}\right)^{\erf(t)}dt\:,$$ the closed form you are looking for is $Z_{\text{era}}(\infty)$ : $$\int_0^\infty \left(e^{-t^2}\right)^{\erf(t)}dt=Z_{\text{era}}(\infty)\simeq 0.972106992769179...$$ Note : The special function $Z_{\text{era}}(x)$ is not standard. As far as I now, no bibliographic reference is available yet. $\endgroup$ – JJacquelin Feb 10 '18 at 10:06
  • $\begingroup$ what do you meant by z_era(x) ? $\endgroup$ – zeraoulia rafik Feb 10 '18 at 10:51
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    $\begingroup$ If you want the closed form for the integral with a variable limit: $$I(a)=\int_{0}^a {(e^{-x²})}^{\erf(x)}dx$$ you should edit the question accordingly. So far your question is asking about the constant $\endgroup$ – Yuriy S Feb 10 '18 at 12:02
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$\newcommand{\erf} {\operatorname{erf}}$I hope this will be helpful to you.

First, abandon the search for closed form. It's not likely it exists.

Now, if you really need a simple expression for $I(a)$ in some range of values, there are ways to get various approximations.

The function is very nice. It goes to its limit at $\infty$ very very fast. Here's the plot of $I(a)$ for $a \in [0,10]$:

enter image description here

So (depending on the accuracy you need) you can easily take $I(a)=I(\infty)$ for $a > a_0$ with $a_0$ around $3$ or $4$.

Mathematica gives for the first $100$ digits:

$$I(\infty)=0.972106992769178593151077875442391175554272\\1833855699009722910408441888759958220033410678218401258734$$


Now, what to do for small $a$?

The function is so nice, you can just use the Taylor expansion around $a =0$. The first term is:

$$I(a) \approx a$$

Here's the plot for $a \in [0,1]$:

enter image description here

The proof is simple. The Taylor series look like this:

$$I(a)=I(0)+I'(0) a+\frac{I''(0)}{2!} a^2+\frac{I'''(0)}{3!} a^3+\dots$$

We can see that:

$$I(0)=0$$

$$I'(0)=e^{-a^2 \erf (a) } \bigg| _{a=0}=1$$

Now let's find a better approximation by computing the higher derivatives:

$$I''(a)=\left( e^{-a^2 \erf (a) } \right)'=-\frac{2}{\sqrt{\pi}} a e^{-a^2 (\text{erf}(a)+1)} \left(\sqrt{\pi } e^{a^2} \text{erf}(a)+a\right)$$

$$I''(0)=0$$


I use Mathematica as a shortcut, but it's easy to do it by hand, if you remember:

$$\erf' (x)=\frac{2}{\sqrt{\pi}} e^{-x^2}$$


$$I'''(0)=0$$

$$I^{ IV} (0)=-\frac{12}{\sqrt{\pi}}$$


So our next approximation is:

$$I(a) \approx a-\frac{ 1}{2\sqrt{\pi}} a^4$$

The plot with both approximations (orange, green) and the function itself (blue) is given below:

enter image description here

You can continue in the same way for higher derivatives.


Now I admit that it's possible you need the values of $I(a)$ for all the possible $a$ and with high precision, so the approximations won't do. Then you need to turn to numerical integration (as Mathematica did for me to plot the function).


Another way to approximate the function is using its derivative:

$$\frac{d I}{da}=e^{-a^2 \erf (a) }$$

But this is an ordinary differential equation, which can be solved numerically.

As an illustration, here's a simple explicit Euler scheme for the step size $h$:

$$\frac{I(a+h)-I(a)}{h}=e^{-a^2 \erf (a) }$$

$$I(a+h)=I(a)+h e^{-a^2 \erf (a) }$$

We can use an initial value $I(0)=0$.

For $h=\frac{1}{10}$ we have the following result (red dots) compared to the exact function (blue line):

enter image description here

For $h=\frac{1}{50}$:

enter image description here

This way can serve as a good alternative to numerical integration (depending on the context and the application of course).

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  • $\begingroup$ Thanks for this attractive answer , really i have tried the approximation with small a i find I(a) almost is a $\endgroup$ – zeraoulia rafik Feb 10 '18 at 13:19
  • $\begingroup$ I should also guess about it's approximation in elemenatry function $\endgroup$ – zeraoulia rafik Feb 10 '18 at 13:42
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    $\begingroup$ @zeraouliarafik, then maybe you can modify your question (adding this information) then you would be able to get a more suitable answer. For what it's worth, you can see that this function behaves more like the error function $\endgroup$ – Yuriy S Feb 10 '18 at 14:16
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    $\begingroup$ I am afraid, I made a mistake in my above comment : The $Z_{\text{era}}(x)$ function better should be the $Y_{\text{uriy}}(x)$ function. $\endgroup$ – JJacquelin Feb 10 '18 at 21:19
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    $\begingroup$ @zeraoulia rafik : Of course this was a joke. I hope you appreciate it. $\endgroup$ – JJacquelin Feb 10 '18 at 21:41

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