Question Number 180350 by mnjuly1970 last updated on 10/Nov/22
Answered by Ar Brandon last updated on 11/Nov/22
$$\mathcal{L}\left({e}^{−{x}} .{erf}\left(\sqrt{{x}}\right)\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {erf}\left(\sqrt{{x}}\right){e}^{−{px}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right){e}^{−\left({p}+\mathrm{1}\right){x}} {dx}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right){e}^{−\left({p}+\mathrm{1}\right){x}} {dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}}\\{\mathrm{v}'\left({x}\right)={e}^{−\left({p}+\mathrm{1}\right){x}} }\end{cases}\:\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)={e}^{−{x}} }\\{\mathrm{v}\left({x}\right)=−\frac{\mathrm{1}}{{p}+\mathrm{1}}{e}^{−\left({p}+\mathrm{1}\right){x}} }\end{cases} \\ $$$$=−\frac{\mathrm{2}}{\:\sqrt{\pi}}\centerdot\frac{\mathrm{1}}{{p}+\mathrm{1}}\left[{e}^{−\left({p}+\mathrm{1}\right){x}} \int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{2}}{\left({p}+\mathrm{1}\right)\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({p}+\mathrm{2}\right){x}} {dx} \\ $$$$=\frac{\mathrm{2}}{\:\left({p}+\mathrm{1}\right)\sqrt{\pi}}+\frac{\mathrm{2}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\sqrt{\pi}}=\frac{\mathrm{2}{p}+\mathrm{6}}{\left({p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{2}\right)\sqrt{\pi}} \\ $$