Question Number 172709 by fabyfb last updated on 30/Jun/22
Answered by Mathspace last updated on 30/Jun/22
$${L}\left({f}\left({x}\right)\right)=\int_{\mathrm{0}} ^{\infty} \:\:{f}\left({t}\right){e}^{−{xt}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{xt}} }{{t}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({t}^{\mathrm{2}} +{a}^{\mathrm{2}} \right){u}} {du}\right){e}^{−{xt}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ut}^{\mathrm{2}} −{xt}} {dt}\right){e}^{−{a}^{\mathrm{2}} {u}} {du} \\ $$$${but}\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ut}^{\mathrm{2}} −{xt}} {dt}=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}\left({t}^{\mathrm{2}} +\mathrm{2}{t}.\frac{{x}}{{u}}+\frac{{x}^{\mathrm{2}} }{{u}^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} }{{u}^{\mathrm{2}} }\right)} \:{dt} \\ $$$$={e}^{\frac{{x}^{\mathrm{2}} }{{u}}} \int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}\left({t}+\frac{{x}}{{u}}\right)^{\mathrm{2}} } \:{dt}\:\:\left({t}+\frac{{x}}{{u}}={z}\right) \\ $$$$={e}^{\frac{{x}^{\mathrm{2}} }{{u}}} \int_{\frac{{x}}{{u}}} ^{+\infty} {e}^{−{uz}^{\mathrm{2}} } {dz}\:\:\:\left(\sqrt{{u}}{z}=\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{u}}}{e}^{\frac{{x}^{\mathrm{2}} }{{u}}} \int_{\frac{{x}}{\:\sqrt{{u}}}} ^{\infty} {e}^{−\alpha^{\mathrm{2}} } \:{d}\alpha \\ $$$$=\frac{\lambda_{\mathrm{0}} }{\:\sqrt{{u}}}{e}^{\frac{{x}^{\mathrm{2}} }{{u}}} \:\:{erf}\left(\frac{{x}}{\:\sqrt{{u}}}\right)\:\Rightarrow \\ $$$${L}\left({f}\left({x}\right)\right)=\lambda_{{o}} \int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{\frac{{x}^{\mathrm{2}} }{{u}}} }{\:\sqrt{{u}}}{erf}\left(\frac{{x}}{\:\sqrt{{u}}}\right)\right){e}^{−{a}^{\mathrm{2}} {u}} {du} \\ $$$$=\lambda_{\mathrm{0}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\:\sqrt{{u}}}{erf}\left(\frac{{x}}{\:\sqrt{{u}}}\right){e}^{−{a}^{\mathrm{2}} {u}+\frac{{x}^{\mathrm{2}} }{{u}}} {du} \\ $$