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

Question

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 ?

Answer 1

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).