Prove the inequality : $\int_{-2}^{2}e!^{-y^{2}}dy<\int_{-2}^{2}e^{-1}dy$

Question

I want to show the following:

$$\int_{-2}^{2}e!^{-y^{2}}dy<\int_{-2}^{2}e^{-1}dy$$

I already know that the left hand side is equal to :

$$\sqrt{\frac{\pi}{\ln\left(e!\right)}}\operatorname{erf}\left(2\sqrt{\ln\left(e!\right)}\right)$$

And the right hand side is equal to :

$$4/e$$

We can also introduce the function :

$$f\left(x\right)=\sqrt{\frac{\pi}{\ln\left(x!\right)+(x-e)^2}}\operatorname{erf}\left(2\sqrt{\ln\left(x!\right)}\right)-\frac{4e^{\frac{1}{9}}}{x^{\left(1+\frac{1}{9}\right)}}$$

And graphically speaking the function $f(x)$ is negative on $x\in[1,\infty)$

I go further :

Using the formula find on wiki page of the Gamma function we have :

$$g\left(x\right)=\left(x+1-\frac{1}{2}\right)\ln\left(x+1\right)-x-1+\frac{1}{2}\ln\left(2\pi\right)+\frac{1}{12\left(x+1\right)}-\frac{1}{360\left(x+1\right)^{3}}\simeq \ln\left(x!\right)$$

So remains to show for $x\ge 1$ :

$$\sqrt{\frac{\pi}{g\left(x\right)+\left(x-e\right)^{2}}}\cdot\operatorname{erf}\left(2\sqrt{g\left(x\right)}\right)-\frac{4e^{\frac{1}{9}}}{x^{\left(1+\frac{1}{9}\right)}}<0$$

For the error function we have also an expansion find on wiki page but it seems to need too much terms to be practicable .

I'm stuck here

Edit 21/08/2022 :

I wasn’t thinking to that, but a simple approximation for the error function is :

$$\operatorname{erf}(x)\simeq \left(\tanh\left(x^{\frac{4}{3}}\right)\right)^{\frac{2}{3}}$$

So my question is how to compare it and show the inequality by hand without a calculator?

Answer 1

The goal is to solve:

$$\int_{-a}^a {e!}^{-y^2}dy=\int_{-a}^a \frac{dy}e\iff\frac{\sqrt\pi\text{erf}\left(a\sqrt{\ln(e!)}\right)}{\sqrt{\ln(e!)}}=\frac2ea$$

Now approximate by using the $21/08/2022$ edit:

$$\begin{align}\frac{\sqrt\pi\text{erf}\left(a\sqrt{\ln(e!)}\right)}{\sqrt{\ln(e!)}}\approx \frac{\sqrt\pi\tanh^\frac23\left(a^\frac43\ln^\frac23(e!)\right)}{\sqrt{\ln(e!)}} =\frac2ea \iff e^{2\ln^\frac23(e!)a^\frac43}=\frac1{\frac12-a^\frac32\frac{\sqrt2\left(\frac{\ln(e!)}{\pi}\right)^\frac34}{e^\frac34}}-1\end{align}$$

A change of variable and your favorite numerical algorithm for finding radicals:

$$a=\sqrt[4]{\left(\frac b{2\ln^\frac23(e!)}\right)^3}:e^b=\frac1{\frac12-\frac{b^\frac98}{2^\frac58e^\frac32\pi^\frac34}}= \frac1{\frac12-\frac{b^\frac98}c},c=16.3097… $$

Now use a numerical method, like approximation via a series about $a=6$, to estimate $e!$ using @Dan’s comment and get:

$$e^6=\frac1{\frac12-\frac{b^\frac98}c}\implies a\approx \frac e2\left(\frac{e^6-3}{e^6-1}\right)^\frac23\sqrt{\frac\pi{\ln(e!)}}=1.9943\cdots$$

Answer 2

All done by hand except the last line.

Starting from @Tyma Gaidash' s answer$$\frac{\sqrt\pi \,\text{erf}\left(a\sqrt{\log(e!)}\right)}{\sqrt{\log(e!)}}=\frac2ea$$ let $x=a\sqrt{\ln(e!)}$ and we need to solve $$e \sqrt{\pi }\, \text{erf}(x)=2 x$$ Square both sides and use $$\text{erf}(x)^2\sim 1-e^{-k x^2} \qquad \text{where} \qquad k=(1+\pi )^{2/3} \log ^2(2)$$ (have a look here for the approximation).

This gives as an approximation $$a=\sqrt{\frac{\frac{e^2 \pi }{4}+\frac 1k W\left(-\frac{k\pi}{4} e^{2-\frac{1}{4} e^2 \pi k} \right)}{\log(e!) }}$$ Numerically, $$a=2.000191\cdots$$