Question Number 200801 by mnjuly1970 last updated on 23/Nov/23
Answered by witcher3 last updated on 24/Nov/23
![introduce erfc(x)=(2/( (√π)))∫_0 ^x e^(−t^2 ) dt φ=∫_0 ^∞ ∫_0 ^∞ ∫_0 ^∞ e^(−(x+y+z)^2 ) dxdydz ∫_0 ^∞ e^(−(x+y+z)^2 ) dx,x+y+z=r =∫_(y+z) ^∞ e^(−r^2 ) dr=((√π)/2)−∫_0 ^(y+z) e^(−r^2 ) dr=((√π)/2)(1−erfc(y+z) φ=((√π)/2)∫_0 ^∞ ∫_0 ^∞ (1−erfc(y+z))dydz=((√π)/2)∫_0 ^∞ ∫_z ^∞ (1−erfc(y))dydz ∫_z ^∞ (1−erfc(y)dy=lim_(x→∞) ∫_z ^x (1−erfc(y))dy =lim_(x→∞) [y−yerfc(y)]_z ^x +(2/( (√π)))∫_z ^x ye^(−y^2 ) dy =lim_(x→∞) (x−xerfc(x)−(e^(−x^2 ) /( (√π)))+zerfc(z)−z+(e^(−z^2 ) /( (√π)))) lim_(x→∞) (x−xerfc(x))=lim_(x→∞) (x−((2x)/( (√π)))∫_0 ^x e^(−t^2 ) dt) =lim_(x→∞) ∣x−x(1−(2/( (√π)))∫_x ^∞ e^(−t^2 ) dt)∣≤lim_(x→∞) ∣((2x)/( (√π)))∫_x ^∞ e^(−t) dt∣ =lim_(x→∞) (xe^(−x) )=0 ∅=((√π)/2)∫_0 ^∞ (zerfc(z)−z+(e^(−z^2 ) /( (√π))))dz ∫_0 ^∞ e^(−z^2 ) =((√π)/2) ∫_0 ^∞ z(erfc(z)−1)dz=[(z^2 /2)(erfc(z)−1)]_0 ^∞ −∫_0 ^∞ (z^2 /2)((2/( (√π)))e^(−z^2 ) )dz =−(1/( (√π)))∫_0 ^∞ z^2 e^(−z^2 ) dz=−(1/(2(√π)))∫_0 ^∞ t^((3/2)−1) e^(−t) =−(1/4) φ=((√π)/2)(−(1/4)+(1/2))=((√π)/8) lim_(x→∞) (∣(x^2 /2)(erfc(x)−1)∣)=(x^2 /2)∫_x ^∞ e^(−t^2 ) ≤(x^2 /2)e^(−x) →0 φ=((√π)/8)](https://www.tinkutara.com/question/Q200813.png)
$$\mathrm{introduce}\:\mathrm{erfc}\left(\mathrm{x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$\phi=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} } \mathrm{dxdydz} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} } \mathrm{dx},\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{r} \\ $$$$=\int_{\mathrm{y}+\mathrm{z}} ^{\infty} \mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr}=\frac{\sqrt{\pi}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{y}+\mathrm{z}} \mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{dr}=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{erfc}\left(\mathrm{y}+{z}\right)\right. \\ $$$$\phi=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\mathrm{erfc}\left(\mathrm{y}+\mathrm{z}\right)\right)\mathrm{dydz}=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{z}} ^{\infty} \left(\mathrm{1}−\mathrm{erfc}\left(\mathrm{y}\right)\right)\mathrm{dydz} \\ $$$$\int_{\mathrm{z}} ^{\infty} \left(\mathrm{1}−\mathrm{erfc}\left(\mathrm{y}\right)\mathrm{dy}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{z}} ^{\mathrm{x}} \left(\mathrm{1}−\mathrm{erfc}\left(\mathrm{y}\right)\right)\mathrm{dy}\right. \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{y}−\mathrm{yerfc}\left(\mathrm{y}\right)\right]_{\mathrm{z}} ^{\mathrm{x}} +\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{z}} ^{\mathrm{x}} \mathrm{ye}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{x}−\mathrm{xerfc}\left(\mathrm{x}\right)−\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\:\sqrt{\pi}}+\mathrm{zerfc}\left(\mathrm{z}\right)−\mathrm{z}+\frac{\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\:\sqrt{\pi}}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{x}−\mathrm{xerfc}\left(\mathrm{x}\right)\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{x}−\frac{\mathrm{2x}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\mid\mathrm{x}−\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{x}} ^{\infty} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\right)\mid\leqslant\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{\mathrm{2x}}{\:\sqrt{\pi}}\int_{\mathrm{x}} ^{\infty} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}\mid \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{xe}^{−\mathrm{x}} \right)=\mathrm{0} \\ $$$$\emptyset=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{zerfc}\left(\mathrm{z}\right)−\mathrm{z}+\frac{\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\:\sqrt{\pi}}\right)\mathrm{dz} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } =\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{z}\left(\mathrm{erfc}\left(\mathrm{z}\right)−\mathrm{1}\right)\mathrm{dz}=\left[\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{erfc}\left(\mathrm{z}\right)−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \right)\mathrm{dz} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} \mathrm{z}^{\mathrm{2}} \mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \mathrm{dz}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} =−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\phi=\frac{\sqrt{\pi}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{8}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mid\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{erfc}\left(\mathrm{x}\right)−\mathrm{1}\right)\mid\right)=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{x}} ^{\infty} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \leqslant\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{e}^{−\mathrm{x}} \rightarrow\mathrm{0} \\ $$$$\phi=\frac{\sqrt{\pi}}{\mathrm{8}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 24/Nov/23
$$\:\:\:\:\:\:{thanks}\:{alot}\:{sir}\:\:{so}\:{nice}\:{solution} \\ $$
Commented by witcher3 last updated on 24/Nov/23
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$