Asymptotics of error function integral with square root

Question

I am interested in the asymptotics of the integral $$I(a):=\int_0^\infty \sqrt{x}\operatorname{Erfc}(x+a)\,\mathrm{d}x$$ for $a>0$. I think that $I(a)$ should be decaying exponentially as $I(a)\lesssim e^{-a^2}$ for large $a$. Numeric integration indicates that $$ \lim_{a\to\infty} I(a) e^{a^2} a^{5/2} =C \in(0,1)$$ for some constant $C$.

Are there integral tables where I could obtain explicit expressions for $I(a)$ for finite $a$? If not, are the asymptotics above correct, and what is $C$?

So far I could only find results when the error function is integrated against integer powers of $x$ e.g. in Section 2.14 of https://intra.ece.ucr.edu/~korotkov/papers/Korotkov-book-integrals.pdf

Answer 1

$$\int_0^\infty \sqrt{x}\,\mathrm{Erfc}(x+a)\,\mathrm{d}x\rightarrow \frac{e^{-a^2}}{(2a)^{5/2}},\;\;\text{for}\;\;a\rightarrow \infty,$$ so $C=2^{-5/2}.$ Corrections are smaller by a factor $1/a$.
I obtained this asymptotics by integrating the large-$a$ expansion of the error-function, $$\text{Erfc}\,(x+a)\rightarrow \frac{e^{-(a+x)^2}}{\sqrt{\pi } a},\;\;\text{for}\;\;a\rightarrow \infty.$$

Answer 2

Too late for an interesting integral.

Using $$e^{(a+x)^2}\, \text{erfc}(x+a)=\frac{1}{\sqrt{\pi }}\sum_{n=1}^\infty (-1)^{n+1}\, P_n(x)\, a^{-n}$$ where the first polynomials are $$\left( \begin{array}{cc} n & P_n(x) \\ 1 & 1 \\ 2 & x \\ 3 & x^2-\frac{1}{2} \\ 4 & x^3-\frac{3 }{2}x \\ 5 & x^4-3 x^2+\frac{3}{4} \\ 6 & x^5-5 x^3+\frac{15} {4}x \\ \end{array} \right)$$

Using the above terms, the definite integral is given by $$I=\frac{e^{-\frac{a^2}{2}}}{64 \sqrt{2 \pi }\,\, a^{11/2}}\left(A(a) K_{\frac{3}{4}}\left(\frac{a^2}{2}\right)-B(a) K_{\frac{1}{4}}\left(\frac{a^2}{2}\right)\right)$$ with $$A(a)=4 a^2 \left(48 a^4+20 a^2-5\right)$$ $$B(a)=192 a^6+160 a^4-20 a^2+5$$ which is a more than decent approaximation (relative error smaller than $0.010$% as soon as $a>4.7$ and smaller than $0.001$% as soon as $a>7.2$).

Expanded again $$I=\frac{e^{-a^2}}{4 \sqrt{2}\, a^{5/2}}\left( 1-\frac{35}{16 a^2}+\frac{3465}{512 a^4}+O\left(\frac{1}{a^6}\right)\right)$$