# $\int\limits_{0}^{t}(\exp{(\sqrt{x}-x^2)} )^ \text{erf(x)}\ \text{d}x,$ VS golden ratio For $t\geq 2$ ,

I have tried to get aother integral representation for the Golden ratio i have got the following representation . This integral is defined as : $$I(t)=\int\limits_{0}^{t}(\exp{(\sqrt{x}-x^2)} )^ \text{erf(x)}\ \text{d}x,$$ and this gives $1.618...$ for $t \to +\infty$ with approximation of $10^{-3}$ as shown here in wolfram alpha and it's seems that is closed nicely to Golden ratio for $t \in [2,3]$ for instance we have $t=2.75..$ and also for $t=e$ , Now my question here is :

Question: Is it possible to say that :For $t\geq 2$ ,$\int\limits_{0}^{t}(\exp{(\sqrt{x}-x^2)} )^ \text{erf(x)}\ \text{d}x$ could be another integral representation of Golden ratio ?

• t=2.7495392638089838109896679573104092694836858310585 with approximation of $\frac{1}{10^{50}}$ – Mariusz Iwaniuk Mar 20 '18 at 16:35
• Thanks for this approximation – zeraoulia rafik Mar 20 '18 at 16:38
• Write $x^2$ instead of $x²$ – Antonio Vargas Mar 21 '18 at 21:11
• t seems pretty arbitrary. – Simply Beautiful Art Mar 29 '18 at 22:48
• For $t=+\infty$ Mma evaluates the integral as $1.6185270961489910965842\ldots$ which is $0.000493107399\ldots$ more than the Golden ratio. Seems like not much of an approximation. – Andrew Mar 31 '18 at 20:38