Question Number 222881 by MrGaster last updated on 10/Jul/25
$$ \\ $$$$\:\int_{\mathrm{0}} ^{\infty} {t}^{{a}} {e}^{−{t}} \mathrm{erf}\left({kt}\right){dt},{a}>\mathrm{0},{k}>\mathrm{0} \\ $$
Answered by gabthemathguy25 last updated on 10/Jul/25
$$\frac{\Gamma\left({a}+\mathrm{1}\right)}{\:\sqrt{\pi}}\centerdot_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\mathrm{1};\frac{\mathrm{3}}{\mathrm{2}}−{k}^{\mathrm{2}} \right) \\ $$
Commented by MrGaster last updated on 10/Jul/25
So the specific proof process ...
Answered by MrGaster last updated on 10/Jul/25
$$ \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{k}} \int_{\mathrm{0}} ^{\infty} {t}^{{a}+\mathrm{1}} {e}^{−{v}^{\mathrm{2}} {t}^{\mathrm{2}} } {dtdv} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{k}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {v}^{\mathrm{2}{n}} }{{n}!}\int_{\mathrm{0}} ^{\infty} {t}^{{a}+\mathrm{1}+\mathrm{2}{n}} {e}^{−{t}} {dtdv} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{k}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {v}^{\mathrm{2}{n}} }{{n}!}\Gamma\left({a}+\mathrm{2}{n}+\mathrm{2}\right){dv} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \Gamma\left({a}+\mathrm{2}{n}+\mathrm{2}\right)}{{n}!}\int_{\mathrm{0}} ^{{k}} {v}^{\mathrm{2}{n}} {dv} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \Gamma\left({a}+\mathrm{2}{n}+\mathrm{2}\right)}{{n}!}\:\frac{{k}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\Gamma\left({a}+\mathrm{1}\right)}{\:\sqrt{\pi}}\centerdot_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\mathrm{1};\frac{\mathrm{3}}{\mathrm{2}};−{k}^{\mathrm{2}} \right) \\ $$