Question Number 197431 by mnjuly1970 last updated on 17/Sep/23
$$ \\ $$$$ \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \left(\:\mathrm{1}+\:{x}^{\mathrm{2}} \:+\:{y}^{\:\mathrm{2}} +{z}^{\:\mathrm{2}} \right)^{\:−\frac{\mathrm{5}}{\mathrm{2}}} {dxdydz}=? \\ $$$$ \\ $$
Answered by witcher3 last updated on 17/Sep/23
![x=rsin(a)cos(b) y=rsin(a)sin(b) y=rcos(a) a∈[0,(π/2)],b∈[0,(π/2)],r∈[0,∞[ dx.dydz=r^2 sin(a)drdadb I=∫_0 ^(π/2) ∫_0 ^(π/2) ∫_0 ^∞ ((r^2 sin(a)drdadb)/((1+r^2 )^(5/2) )) =(π/2)∫_0 ^(π/2) sin(a)da∫_0 ^∞ (r^2 /((1+r^2 )^(5/2) ))dr r^2 =y⇒dr=(1/2)y^(−(1/2)) dy ∫_0 ^∞ (r^2 /((1+r^2 )^(5/2) ))dr=∫_0 ^∞ (y^(1/2) /(2(1+y)^(5/2) ))dy =(1/2)β((3/2),1)=((Γ((3/2)))/(2Γ((5/2))))=(1/3) ∫_0 ^(π/2) sin(a)da=1 I=(π/2).1.(1/3)=(π/6)](https://www.tinkutara.com/question/Q197432.png)
$$\mathrm{x}=\mathrm{rsin}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right) \\ $$$$\mathrm{y}=\mathrm{rsin}\left(\mathrm{a}\right)\mathrm{sin}\left(\mathrm{b}\right) \\ $$$$\mathrm{y}=\mathrm{rcos}\left(\mathrm{a}\right) \\ $$$$\mathrm{a}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right],\mathrm{b}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right],\mathrm{r}\in\left[\mathrm{0},\infty\left[\right.\right. \\ $$$$\mathrm{dx}.\mathrm{dydz}=\mathrm{r}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{a}\right)\mathrm{drdadb} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{r}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{a}\right)\mathrm{drdadb}}{\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\left(\mathrm{a}\right)\mathrm{da}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} }\mathrm{dr} \\ $$$$\mathrm{r}^{\mathrm{2}} =\mathrm{y}\Rightarrow\mathrm{dr}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dy} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{r}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} }\mathrm{dr}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{1}+\mathrm{y}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\left(\mathrm{a}\right)\mathrm{da}=\mathrm{1} \\ $$$$\mathrm{I}=\frac{\pi}{\mathrm{2}}.\mathrm{1}.\frac{\mathrm{1}}{\mathrm{3}}=\frac{\pi}{\mathrm{6}} \\ $$
Commented by Frix last updated on 17/Sep/23
Nice!
Commented by mnjuly1970 last updated on 17/Sep/23
$${thanks}\:{alot}\:{sir}\:{Witcher} \\ $$
Commented by witcher3 last updated on 17/Sep/23
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$
Answered by Frix last updated on 17/Sep/23
![In steps ∫_0 ^∞ (x^2 +a)^(−(5/2)) dx=[((x(2x^2 +3a))/(3a^2 (x^2 +a)^(3/2) ))]_0 ^∞ =(2/(3a^2 )) (2/3)∫_0 ^∞ (y^2 +b)^(−2) dy=[(y/(3b(y^2 +b)))−((tan^(−1) (y/( (√b))))/(3b^(3/2) ))]_0 ^∞ =(π/(6b^(3/2) )) (π/6)∫_0 ^∞ (z^2 +1)^(−(3/2)) dz=[((πz)/(6(√(z^2 +1))))]_0 ^∞ =(π/6)](https://www.tinkutara.com/question/Q197433.png)
$$\mathrm{In}\:\mathrm{steps} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({x}^{\mathrm{2}} +{a}\right)^{−\frac{\mathrm{5}}{\mathrm{2}}} {dx}=\left[\frac{{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{a}\right)}{\mathrm{3}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{2}}{\mathrm{3}{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({y}^{\mathrm{2}} +{b}\right)^{−\mathrm{2}} {dy}=\left[\frac{{y}}{\mathrm{3}{b}\left({y}^{\mathrm{2}} +{b}\right)}−\frac{\mathrm{tan}^{−\mathrm{1}} \:\frac{{y}}{\:\sqrt{{b}}}}{\mathrm{3}{b}^{\frac{\mathrm{3}}{\mathrm{2}}} }\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{6}{b}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\frac{\pi}{\mathrm{6}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dz}=\left[\frac{\pi{z}}{\mathrm{6}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{6}} \\ $$