Question Number 180350 by mnjuly1970 last updated on 10/Nov/22
Answered by Ar Brandon last updated on 11/Nov/22
![L(e^(−x) .erf((√x)))=∫_0 ^∞ e^(−x) erf((√x))e^(−px) dx =∫_0 ^∞ ((2/( (√π)))∫_0 ^(√x) e^(−t^2 ) dt)e^(−(p+1)x) dx=(2/( (√π)))∫_0 ^∞ (∫_0 ^(√x) e^(−t^2 ) dt)e^(−(p+1)x) dx { ((u(x)=∫_0 ^(√x) e^(−t^2 ) dt)),((v′(x)=e^(−(p+1)x) )) :} ⇒ { ((u′(x)=e^(−x) )),((v(x)=−(1/(p+1))e^(−(p+1)x) )) :} =−(2/( (√π)))∙(1/(p+1))[e^(−(p+1)x) ∫_0 ^(√x) e^(−t^2 ) dt]_0 ^∞ +(2/((p+1)(√π)))∫_0 ^∞ e^(−(p+2)x) dx =(2/( (p+1)(√π)))+(2/((p+1)(p+2)(√π)))=((2p+6)/((p^2 +3p+2)(√π)))](Q180372.png)
$$\mathcal{L}\left({e}^{−{x}} .{erf}\left(\sqrt{{x}}\right)\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {erf}\left(\sqrt{{x}}\right){e}^{−{px}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right){e}^{−\left({p}+\mathrm{1}\right){x}} {dx}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right){e}^{−\left({p}+\mathrm{1}\right){x}} {dx} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}}\\{\mathrm{v}'\left({x}\right)={e}^{−\left({p}+\mathrm{1}\right){x}} }\end{cases}\:\Rightarrow\begin{cases}{\mathrm{u}'\left({x}\right)={e}^{−{x}} }\\{\mathrm{v}\left({x}\right)=−\frac{\mathrm{1}}{{p}+\mathrm{1}}{e}^{−\left({p}+\mathrm{1}\right){x}} }\end{cases} \\ $$$$=−\frac{\mathrm{2}}{\:\sqrt{\pi}}\centerdot\frac{\mathrm{1}}{{p}+\mathrm{1}}\left[{e}^{−\left({p}+\mathrm{1}\right){x}} \int_{\mathrm{0}} ^{\sqrt{{x}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{2}}{\left({p}+\mathrm{1}\right)\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({p}+\mathrm{2}\right){x}} {dx} \\ $$$$=\frac{\mathrm{2}}{\:\left({p}+\mathrm{1}\right)\sqrt{\pi}}+\frac{\mathrm{2}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\sqrt{\pi}}=\frac{\mathrm{2}{p}+\mathrm{6}}{\left({p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{2}\right)\sqrt{\pi}} \\ $$