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 bycandre last updated on 13/May/18
![u=x^(n−1) ⇒du=(n−1)x^(n−2) dx dv=xe^(−(x^2 /p)) dx⇒v=∫xe^(−(x^2 /p)) dx w=−(x^2 /p)⇒dw=−((2x)/p)dx v=−(p/2)∫e^w dw=−(p/2)e^(−(x^2 /p)) ∫_0 ^∞ x^n e^(−(x^2 /p)) dx=−[((px^(n−1) e^(−(x^2 /p)) )/2)]_0 ^∞ +(((n−1)p)/2)∫_0 ^∞ x^(n−2) e^(−(x^2 /p)) dx J_(n,p) =(((n−1)pJ_(n−2,p) )/2)](Q34944.png)
$${u}={x}^{{n}−\mathrm{1}} \Rightarrow{du}=\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {dx} \\ $$ $${dv}={xe}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx}\Rightarrow{v}=\int{xe}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx} \\ $$ $$\:{w}=−\frac{{x}^{\mathrm{2}} }{{p}}\Rightarrow{dw}=−\frac{\mathrm{2}{x}}{{p}}{dx} \\ $$ $${v}=−\frac{{p}}{\mathrm{2}}\int{e}^{{w}} {dw}=−\frac{{p}}{\mathrm{2}}{e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} \\ $$ $$\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{{n}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx}=−\left[\frac{{px}^{{n}−\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} +\frac{\left({n}−\mathrm{1}\right){p}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{{n}−\mathrm{2}} {e}^{−\frac{{x}^{\mathrm{2}} }{{p}}} {dx} \\ $$ $${J}_{{n},{p}} =\frac{\left({n}−\mathrm{1}\right){pJ}_{{n}−\mathrm{2},{p}} }{\mathrm{2}} \\ $$