Integral of product of Gamma, Erfc and Exponential

Question

I can confirm that: $$\int_{-\infty }^{\infty } \frac{e^{-\frac{K^2}{2}} 2^{-n-\frac{1}{2}} n \Gamma \left(\frac{1}{2},\frac{K^2}{2}\right) \text{erfc}\left(\frac{-K}{\sqrt{2}}\right)^{n-1}}{\pi } \, dK = \frac{1-2^{-n}}{n+1}$$ numerically, but I can't work out the derivation symbolically. Is there a symbolic derivation?

Answer 1

Substitute $x=\operatorname{erfc}(-K/\sqrt2)$; then $dx/dK=\sqrt{2/\pi}e^{-K^2/2}$, and $\color{blue}{\Gamma(1/2,z)=\sqrt\pi\operatorname{erfc}\sqrt{z}}$ for $z\geqslant 0$, thus $\Gamma(1/2,K^2/2)=\sqrt\pi\operatorname{erfc}(|K|/\sqrt2)=\sqrt\pi(1-|1-x|)$. The given integral is $$2^{-n-1/2}\frac{n}{\pi}\int_0^2\sqrt\pi(1-|1-x|)x^{n-1}\sqrt{\pi/2}\,dx=\frac{n}{2^{n+1}}\int_0^2 x^{n-1}(1-|1-x|)\,dx.$$ This is evaluated easily (say, by splitting $\int_0^2=\int_0^1+\int_1^2$), and the result is as expected.