Question Number 133454 by rs4089 last updated on 22/Feb/21
Answered by TheSupreme last updated on 22/Feb/21
![Ω={(x,y)∣ x>0 ; 0<y<x} Ω={(x,y)∣ y>0, x>y} I=∫_0 ^∞ ∫_0 ^x e^(−xy) ydydx = ∫^∞ _0 ∫^∞ _y e^(−xy) ydydx ∫_0 ^∞ ∫_y ^∞ e^(−xy) ydxdy=∫_0 ^∞ [−(e^(−xy) /y^2 )]_y ^∞ dy ∫_0 ^∞ (e^(−y^2 ) /y^2 )dy f(x)= e^(−y^2 ) →f′(y)=−2ye^y^2 g′(x)= (1/y^2 ) → g(x)=−(1/y) ∫f(x)g′(x)=f(x)g(x)−∫f′(x)g(x) ∫_0 ^∞ e^(−y^2 ) /y^2 dy=−(e^(−y^2 ) /y)−∫2e^(−y^2 ) dy I=(e^(−y^2 ) /y)−2 ((√π)/2) I=(e^(−y^2 ) /y)−(√π)](Q133459.png)
$$\Omega=\left\{\left({x},{y}\right)\mid\:{x}>\mathrm{0}\:;\:\mathrm{0}<{y}<{x}\right\} \\ $$$$\Omega=\left\{\left({x},{y}\right)\mid\:{y}>\mathrm{0},\:{x}>{y}\right\} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{{x}} {e}^{−{xy}} {ydydx}\:=\:\underset{\mathrm{0}} {\int}^{\infty} \underset{{y}} {\int}^{\infty} {e}^{−{xy}} {ydydx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{{y}} ^{\infty} {e}^{−{xy}} {ydxdy}=\int_{\mathrm{0}} ^{\infty} \left[−\frac{{e}^{−{xy}} }{{y}^{\mathrm{2}} }\right]_{{y}} ^{\infty} {dy} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{y}^{\mathrm{2}} } }{{y}^{\mathrm{2}} }{dy}\: \\ $$$${f}\left({x}\right)=\:{e}^{−{y}^{\mathrm{2}} } \:\:\:\rightarrow{f}'\left({y}\right)=−\mathrm{2}{ye}^{{y}^{\mathrm{2}} } \\ $$$${g}'\left({x}\right)=\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:\:\rightarrow\:{g}\left({x}\right)=−\frac{\mathrm{1}}{{y}} \\ $$$$\int{f}\left({x}\right){g}'\left({x}\right)={f}\left({x}\right){g}\left({x}\right)−\int{f}'\left({x}\right){g}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{y}^{\mathrm{2}} } /{y}^{\mathrm{2}} {dy}=−\frac{{e}^{−{y}^{\mathrm{2}} } }{{y}}−\int\mathrm{2}{e}^{−{y}^{\mathrm{2}} } {dy} \\ $$$${I}=\frac{{e}^{−{y}^{\mathrm{2}} } }{{y}}−\mathrm{2}\:\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$${I}=\frac{{e}^{−{y}^{\mathrm{2}} } }{{y}}−\sqrt{\pi} \\ $$