I am trying to solve the following integral $$\int_0^{\infty } \int_{a-b x}^{\infty } \exp \left(-u^2\right) \, du \, dx.$$ I know it can be represented as an integral of the complementary error function, but I was wondering if there is a closed form.

May be using series expansion ?

Thank You.


1 Answer 1


From definition $$\int \exp \left(-u^2\right) \, du=\frac{ \sqrt{\pi }}{2}\, \text{erf}(u)$$ which makes $$\int_{a-b x}^{\infty } \exp \left(-u^2\right) \, du=\frac{\sqrt{\pi }}{2} \, \text{erfc}(a-b x)$$ So, $$\int \int_{a-b x}^{\infty } \exp \left(-u^2\right) \, du \, dx=\frac{\sqrt{\pi }}{2} \,\int\text{erfc}(a-b x) \,dx=-\frac{\sqrt{\pi } }{2 b}\int\text{erfc}(y)\, dy$$

Now, you can integrate by parts and get $$\int \text{erfc}(y)=y \,\text{erfc}(y)-\frac{1}{\sqrt{\pi }}\,e^{-y^2}$$

Back to $x$, we then have to consider the limit, for $t\to \infty$, of $$-\frac{e^{-a^2}+\sqrt{\pi } (-a \,\text{erf}(a-b t)-b t \,\text{erfc}(a-b t)+a \, \text{erf}(a))-e^{-(a-b t)^2}}{2 b}$$ which does not exist if $b>0$. If $b<0$, the limit should be $$\frac{\sqrt{\pi }\, a\, \text{erfc}(a)-e^{-a^2}}{2 b}$$


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