# Is really this :$\int_{0}^{t}\operatorname{erf}(x+\sqrt{1-\log (x)} )dx \sim t$ true for every $t$?

I have accrossed this integral when I run some of my computation in Wolfram alpha with many values of $$t$$ , Really seems to conjecture that :

$$\int_{0}^{t}\operatorname{erf}(x+\sqrt{1-\log (x)} )dx \sim t$$ for every $$t$$ , The key I dea i have used is the series expansion of that function $$x+\sqrt{1-\log (x)}$$ at $$x=e$$ as shown in the below image , by substitution in the titled function that's make the integral is very complicated and as a reason we do not have a closed form for it this make obstacle to me to get what i have conjuctred , Now my justification if the conjecture is true is that :$$\lim_{x\to \infty}\operatorname{erf}(x+\sqrt{1-\log (x)} ) =\lim_{x\to e}\operatorname{erf}(x+\sqrt{1-\log (x)} )=1$$ , But this is not enough and is not clear at all , Now my question how do i show the titled equality if it is true ? • What's the meaning of this integral for $t \geq e$? – Ian Jul 24 '19 at 22:28
• it's not defined for x greater than e – zeraoulia rafik Jul 24 '19 at 22:29
• You say "for every $t$", that's the objection I am raising. Besides that, what you are really doing is taking an estimate valid for small $x$ which is that for such small $x$ you have $\sqrt{1-\log(x)} \approx +\infty$ so $\mathrm{erf}(x+\sqrt{1-\log(x)}) \approx \lim_{x \to \infty} \mathrm{erf}(x) = 1$ and you get an estimate of $t$ for the integration. As $t \to 0$ this estimate is correct to leading order; for finite $t$ you need to say something about the error in the approximation more carefully (which will be a bit tricky because the inner function is singular). – Ian Jul 24 '19 at 22:30
• For t greater than e Wolfram alpha assumed complex closed values to 0 in imaginary part , I said for every to since the imaginary part go to be 0 , and i havn't took any condition where should i compute that integral over R or C – zeraoulia rafik Jul 24 '19 at 22:34
• $\operatorname{erf}(x+\sqrt{1-\ln(x)})$ will be very close to $1$. The minimum is $\approx 0.988$. This is why the integral is approximately equal to $t$. – Varun Vejalla Jul 24 '19 at 22:42

automaticallyGenerated gave the correct explanation. Just let me elaborate considering $$f(x)=\operatorname{erf}(x+\sqrt{1-\ln(x)})$$ for which $$f'(x)=\frac{2}{\sqrt{\pi }}e^{-\left(x+\sqrt{1-\log (x)}\right)^2} \left(1-\frac{1}{2 x \sqrt{1-\log(x)}}\right)$$ So, the derivative cancels when $$2 x \sqrt{1-\log(x)}=1$$ that is to say when $$1+\frac 12 \log\left(\frac 1 {x^2} \right)=\frac 1 {4x^2}$$ the solution of which being $$x_*=\sqrt{2 W\left(\frac{e^2}{2}\right)}\approx 1.52260$$ where $$W(.)$$ is Lambert function. $$f(x_*)=0.998762$$
Interesting would be the Taylor expansion around $$x=1$$; this would be $$f(x)=\text{erf}(2)+\frac{x-1}{e^4 \sqrt{\pi }}-\frac{3 (x-1)^2}{4 e^4 \sqrt{\pi }}-\frac{(x-1)^3}{8 e^4 \sqrt{\pi }}+\frac{137 (x-1)^4}{192 e^4 \sqrt{\pi }}+O\left((x-1)^5\right)$$ making $$\int_0^t f(x)\,dx\approx\left(\text{erf}(2)-\frac{175}{192 e^4 \sqrt{\pi }}\right) t-\frac{35 t^2}{96 e^4 \sqrt{\pi }}+\frac{125 t^3}{96 e^4 \sqrt{\pi }}-\frac{143 t^4}{192 e^4 \sqrt{\pi }}+\frac{137 t^5}{960 e^4 \sqrt{\pi }}$$ which, numerically is $$0.985904 t-0.00376742 t^2+0.0134551 t^3-0.0076963 t^4+0.00147468 t^5$$