Solving the integral $\int_{0}^{t} ds \sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right)$

Question

I want to solve the following integral $$ \int_{0}^{t} ds \sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right) $$ with $a\in\mathbb{R}$ and $t\in\mathbb{R}^+$. I have tried some substitutions and the most promising are

1. $u=\sqrt{t-s} \ \left(du=-\frac{1}{2\sqrt{t-s}} \ ds\right)$ so that the integral becomes \begin{align} \int_{0}^{t} ds \sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right) &=-2\int_{\sqrt{t}}^{0}du\ \sqrt{s}\operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right)\\ &=\color{orange}{2\int^{\sqrt{t}}_{0}du\ \sqrt{t-u^2}\operatorname{erfc}\left(\frac{a}{\sqrt{t-u^2}}\right).} \end{align}
2. $v=\frac{1}{\sqrt{s}} \ \left(dv = -\frac{1}{2s^{3/2}} \ \ \ ds \right)$ which yields \begin{align} \int_{0}^{t} ds \sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right) &=2\int_{\tfrac{1}{\sqrt{t}}}^{\infty} dv \ v^{3/2} \frac{1}{v\sqrt{t-\tfrac{1}{v^2}}} \ \operatorname{erfc}\left(av\right)\\ &=\color{orange}{2\int_{\tfrac{1}{\sqrt{t}}}^{\infty} dv \ \frac{v^{3/2}}{\sqrt{v^2t-1}} \ \operatorname{erfc}\left(av\right)} \end{align}

Furthermore, I have looked at the table of integrals [1], [2] and [3]. However, so far, no luck. Is there someone who knows a way forward?

Answer 1

Let $s=\frac {a^2}{x^2}$ to make $$\int_{0}^{t}\sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right)\,ds=2a^3\int_{\frac{a}{\sqrt{t}}}^\infty\frac{\text{erfc}(x)}{x^3 \sqrt{t x^2-a^2}}\,dx$$

I do not think that the antiderivative can be computed but the definite integral is

$$I=\int_{0}^{t}\sqrt{\frac{s}{t-s}} \operatorname{erfc}\left(\frac{a}{\sqrt{s}}\right)\,ds=t \,\,G_{2,3}^{3,0}\left(\frac{a^2}{t}| \begin{array}{c} 1,2 \\ 0,\frac{1}{2},\frac{3}{2} \end{array} \right)$$

where appears the Meijer G function. This has been checked numerically.

Let $y=\frac t{a^2}$ and then $$\frac I{a^2}=y \,\,G_{2,3}^{3,0}\left(\frac{1}{y}| \begin{array}{c} 1,2 \\ 0,\frac{1}{2},\frac{3}{2} \end{array} \right)$$

To allow comparison with numerical integration, I give you below a few values $$\left( \begin{array}{cc} y & \frac I{a^2} \\ 0 & 0.00000 \\ 1 & 0.16389 \\ 2 & 0.74897 \\ 3 & 1.55103 \\ 4 & 2.47510 \\ 5 & 3.47872 \\ 6 & 4.53912 \\ 7 & 5.64249 \\ 8 & 6.77981 \\ 9 & 7.94477 \\ 10 & 9.13278 \end{array} \right)$$

Edit

I forgot to mention that this is valid only for $a>0$. I did not find any solution for $a<0$.

What is interesting is the expansion of $\frac I{a^2}$ for large values of $y$. It write $$\frac I{a^2}=\frac \pi 2 y-\frac{4 }{\sqrt{\pi }}\sqrt{y}\Bigg[1+ \sum_{n=1}^\infty (-1)^n \frac {\alpha_n+\beta_n \big[\log(4y)-\gamma\big]} {\delta_n\,y^n}\Bigg]$$ The first $\alpha_n$ make the sequence $$\{5,9,20,193,921,1075,9839,265715,\cdots\}$$ the first $\beta_n$ $$\{3,10,21,180,770,819,6930,175032,\cdots\}$$ the first $\delta_n$ $$\{18,400,4704,248832,7434240,62300160,4644864000,1145485393920,\cdots\}$$ None of them in $OEIS$.

Using this truncated expansion for $y=1$ leads to an absolute error of $7.13\times 10^{-8}$ and, for $y=10$ to an absolute error of $5.33\times 10^{-15}$.

Answer 2

If you substitute $av=x$ you are left with the problem of evaluating an integral of the form $$ \int_z^{ + \infty } {\frac{\operatorname{erfc}x}{{x^2 }}dx},\quad z>0 . $$ Using integration by parts once yields $$ \int_z^{ + \infty } {\frac{{\operatorname{erfc}x}}{{x^2 }}dx} = \frac{{\operatorname{erfc}z}}{z} - \frac{2}{{\sqrt \pi }}\int_z^{ + \infty } {\frac{1}{x}e^{ - x^2 } dx} = \frac{{\operatorname{erfc}z}}{z} - \frac{1}{{\sqrt \pi }}E_1 (z^2 ), $$ where $E_1$ is the exponential integral.

Answer 3

Let us start from the second substitution in the question, i.e. \begin{align} f(a) =2\int_{\tfrac{1}{\sqrt{t}}}^{\infty} dv \ \frac{v^{3/2}}{\sqrt{v^2t-1}} \ \operatorname{erfc}\left(av\right) \end{align} and take the partial derivative w.r.t. the variable a \begin{align} \partial_a f(a) & =2\int_{\tfrac{1}{\sqrt{t}}}^{\infty} dv \ \frac{v^{3/2}}{\sqrt{v^2t-1}} \ \left[ -\frac{2 v e^{-a^2 v^2}}{\sqrt{\pi }} \right] \\ & =-\frac{4}{\sqrt{\pi}}\int_{\tfrac{1}{\sqrt{t}}}^{\infty} dv \ \frac{v^{5/2}}{\sqrt{v^2t-1}}e^{-a^2 v^2} \end{align} Mathematica solves the remaining integral for all $t > 0$ \begin{align} & \partial_a f(a) \\ & =-\frac{4}{\sqrt{\pi}}\left[ \frac{\frac{\ \Gamma\left[\frac{5}{4}\right] \, _1F_1\left[\frac{1}{2},-\frac{1}{4},-\frac{a^{2}}{t}\right]}{\left(\frac{a^{2}}{t}\right)^{5 / 4}}+\frac{\sqrt{\pi} \ \Gamma\left[-\frac{5}{4}\right] \, _1F_1\left[\frac{7}{4}, \frac{9}{4},-\frac{a^{2}}{t}\right]}{\ \Gamma\left[-\frac{3}{4}\right]}}{2 t^{7 / 4}} \right] \\ & =-\frac{4}{\sqrt{\pi}}\left[ \frac{\sqrt{\pi } \ \Gamma \left(-\frac{5}{4}\right) \, _1F_1\left(\frac{7}{4};\frac{9}{4};-\frac{a^2}{t}\right)}{2 t^{7/4} \ \Gamma \left(-\frac{3}{4}\right)}+\frac{\sqrt[4]{t} \ \Gamma \left(\frac{5}{4}\right) \left(\frac{a^2}{t}\right)^{3/4} \, _1F_1\left(\frac{1}{2};-\frac{1}{4};-\frac{a^2}{t}\right)}{2 a^4} \right] \\ & =-\frac{4}{\sqrt{\pi}}\left[ \frac{\sqrt{\pi } }{2} t^{-7/4}\frac{\ \Gamma \left(-\frac{5}{4}\right) }{ \ \Gamma \left(-\frac{3}{4}\right)} \, _1F_1\left(\frac{7}{4};\frac{9}{4};-\frac{a^2}{t}\right)+\frac{1}{2 \sqrt{t} a^{5/2}} \ \Gamma \left(\frac{5}{4}\right) \, _1F_1\left(\frac{1}{2};-\frac{1}{4};-\frac{a^2}{t}\right) \right] \end{align} We know that \begin{align} f(\infty)-f(a) = \int_a^\infty \partial_a f(a) \end{align} Because $f(\infty=0)$ one can write \begin{align} f(a)=- \int_{a}^{\infty} \partial_a f(a) \end{align} From now we must set a constraint. If somebody can solve for a general $a \in \mathbb{R}$, please let me now!

If we assume that $a>0$, Mathematica solves the integral which gives \begin{align} & f(a) \\ & = \frac{4}{\sqrt{\pi}}\left[ \frac{\frac{10 \ t^{5/4} \ \ \ \Gamma \left(\frac{1}{4}\right) \ _2F_2 \ \left(-\frac{3}{4},\frac{1}{2};-\frac{1}{4},\frac{1}{4};-\frac{a^2}{t}\right)}{a^{3/2}} \ \ \ +\sqrt{2} \left(\frac{128 a \ \Gamma \left(\frac{7}{4}\right)^2 \ \ _2F_2 \ \left(\frac{1}{2},\frac{7}{4};\frac{3}{2},\frac{9}{4};-\frac{a^2}{t}\right)}{\sqrt{\pi }} \ \ +5 \sqrt{t} \ \Gamma \left(\frac{1}{4}\right)^2\right)}{120 t^{7/4}} \right] \end{align} Simplifying and putting on a multiline yield \begin{multline} f(a) = \frac{4}{\sqrt{\pi}}\frac{\ \Gamma \left(\frac{1}{4}\right)^2}{12 \sqrt{2} t^{5/4}}+\frac{4}{\sqrt{\pi}}\frac{16 \sqrt{\frac{2}{\pi }} a \ \Gamma \left(\frac{7}{4}\right)^2 \ _2F_2\left(\frac{1}{2},\frac{7}{4};\frac{3}{2},\frac{9}{4};-\frac{a^2}{t}\right)}{15 t^{7/4}}\\ +\frac{4}{\sqrt{\pi}}\frac{\ \Gamma \left(\frac{1}{4}\right) \ _2F_2\left(-\frac{3}{4},\frac{1}{2};-\frac{1}{4},\frac{1}{4};-\frac{a^2}{t}\right)}{12 a^{3/2} \sqrt{t}} \end{multline} Doing this once more gives \begin{multline} f(a) = \frac{1}{3\sqrt{2\pi}t^{5/4}}\ \Gamma \left(\frac{1}{4}\right)^2+\frac{ 64 \sqrt{2}\ a }{\sqrt{\pi}15 t^{7/4}} \ \Gamma \left(\frac{7}{4}\right)^2 \ _2F_2\left(\frac{1}{2},\frac{7}{4};\frac{3}{2},\frac{9}{4};-\frac{a^2}{t}\right)\\ +\frac{1}{ 3\sqrt{\pi}a^{3/2} \sqrt{t}}\ \Gamma \left(\frac{1}{4}\right) \ _2F_2\left(-\frac{3}{4},\frac{1}{2};-\frac{1}{4},\frac{1}{4};-\frac{a^2}{t}\right) \end{multline} $\color{red}{\text{Again, we have assumed that $a>0$ would be nice also find the expression for }a \in \mathbb{R}.}$