Question Number 34910 by abdo imad last updated on 12/May/18
$${find}\:{J}_{{n},{p}} \:=\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:\:{e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} \:\:{dx}\:\:{with}\:{p}>\mathrm{0}\:{and}\:{n}\:{integr} \\ $$
Commented by candre last updated on 13/May/18
$${u}={x}^{{n}−\mathrm{1}} \Rightarrow{du}=\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {dx} \\ $$$${dv}={xe}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx}\Rightarrow{v}=\int{xe}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx} \\ $$$$\:{w}=−\frac{{x}^{\mathrm{2}} }{{p}}\Rightarrow{dw}=−\frac{\mathrm{2}{x}}{{p}}{dx} \\ $$$${v}=−\frac{{p}}{\mathrm{2}}\int{e}^{{w}} {dw}=−\frac{{p}}{\mathrm{2}}{e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{{n}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx}=−\left[\frac{{px}^{{n}−\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} +\frac{\left({n}−\mathrm{1}\right){p}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{{n}−\mathrm{2}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx} \\ $$$${J}_{{n},{p}} =\frac{\left({n}−\mathrm{1}\right){pJ}_{{n}−\mathrm{2},{p}} }{\mathrm{2}} \\ $$