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Can someone help about proving the following relationship:

\begin{equation} \operatorname{erfc}\left(\sqrt a\right) = \int_{0}^{\infty} e^{-a} \times \frac{\sqrt a}{\sqrt \pi} \times \frac{\operatorname{erf}\left(\sqrt y\right)}{ \left(a + y\right)^{2}} \,dy \end{equation}

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Just use the identity $$ \frac{1}{(a+y)^2}=\int_0^\infty d\xi\ \xi e^{-\xi (a+y)} $$ to rewrite the r.h.s. as $$ e^{-a}\sqrt{\frac{a}{\pi}}\int_0^\infty d\xi\ \xi\ e^{-a\xi} \int_0^\infty dy\ \mathrm{erf}(\sqrt{y})e^{-\xi y}\ . $$ The $y$-integral admits an explicit antiderivative $$ \int dy\ \mathrm{erf}(\sqrt{y})e^{-\xi y}=\frac{\text{erf}\left(\sqrt{\xi +1} \sqrt{y}\right)}{\xi \sqrt{\xi +1}}-\frac{\text{erf}\left(\sqrt{y}\right) e^{-\xi y}}{\xi }+C\ , $$ as one can verify by differentiation. Therefore, the definite integral reads $$ \int_0^\infty dy\ \mathrm{erf}(\sqrt{y})e^{-\xi y}=\frac{1}{\xi \sqrt{\xi +1}}\ , $$ so that the r.h.s. is effectively equal to $$ e^{-a}\sqrt{\frac{a}{\pi}}\int_0^\infty d\xi\frac{e^{-a \xi}}{\sqrt{\xi+1}}\ , $$ whose antiderivative is also known $$ \int d\xi\frac{e^{-a \xi}}{\sqrt{\xi+1}}=\frac{\sqrt{\pi } e^a \text{erf}\left(\sqrt{a} \sqrt{\xi +1}\right)}{\sqrt{a}}+C_1. $$ Using the known asymptotics $\text{erf}\left(\sqrt{a} \sqrt{\xi +1}\right)\to 1$ for $\xi\to\infty$, the claimed equality with $\text{erfc}(\sqrt{a})=1-\text{erf}(\sqrt{a})$ is readily established.

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