# How this $\int_0^{a}x^\text{erf(exp(-x))}dx$ behaves?

Really i'm confused about behavior of this integral : $\int_0^{a}x^\text{erf(exp(-x))}dx$ using wolfram alpha for some values it's seems to me that integral satisfying : $I(a)=a-0.14...$ , for $a \geq 10$ , but i didn't succeed to get it closed form since it is divergent , Then I want to know more about it's behavior ?

• Is the integrand strictly positive over the interval of integration? Does it tend to some non-zero value? – Michael McGovern Apr 22 '18 at 1:36

The leading behavior as $a\to\infty$ is easy. Just define $$I(a)=\int_0^{a}x^\text{erf(exp(-x))}dx\ ,$$ and take the derivative $$I'(a)=e^{\log(a)\cdot\mathrm{erf}(e^{-a})}\ .$$ One can show that $$\lim_{a\to\infty} I'(a)=1$$ (which implies the behavior $I(a)\sim a$) by the substitution $e^{-a}=t$: $$\lim_{a\to\infty}e^{\log(a)\cdot\mathrm{erf}(e^{-a})}=\lim_{t\to 0}e^{\log(-\log t)\cdot\mathrm{erf}(t)}$$ and using $\mathrm{erf}(t)\sim 2t/\sqrt{\pi}$ as $t\to 0$.
To compute the subleading contribution, we need to study $$J(a)=\int_0^{a}x^\text{erf(exp(-x))}dx-a=\int_0^{a}(x^\text{erf(exp(-x))}-1)dx\ ,$$ which leads to a convergent integral as $a\to\infty$. Therefore, the sought asymptotics reads $$I(a)\sim a+\int_0^{\infty}(x^\text{erf(exp(-x))}-1)dx+o(1)\ ,\qquad\mbox{for }a\to\infty\ ,$$ where the constant term $$\int_0^{\infty}(x^\text{erf(exp(-x))}-1)dx\approx -0.14113...$$ (there is no hope to find a closed form for this integral, though - but it's fine, it's just a constant).