Question Number 75027 by chess1 last updated on 06/Dec/19
Commented by mathmax by abdo last updated on 06/Dec/19
![x+z=3 ⇒0≤x≤3 and 0≤z≤3 we have 0≤y≤2 ⇒ ∫∫∫ ((dxdydz)/((x+y+z)^3 )) =∫_0 ^3 ( ∫_o ^2 (∫_0 ^3 (dx/((x+y+z)^3 )))dy)dz we have ∫_0 ^3 (dx/((x+y+z)^3 )) =[−(1/2)(x+y+z)^(−2) ]_(x=0) ^3 =−(1/2){(3+y+z)^(−2) −(y+z)^(−2) } =(1/(2(y+z)^2 ))−(1/((y+z+3)^2 )) ⇒ ∫_0 ^2 (∫_0 ^3 (dx/((x+y+z)^3 )))dy =(1/2) ∫_0 ^2 (dy/((y+z)^2 ))−(1/2)∫_0 ^2 (dy/((y+z+3)^2 )) =(1/2)[((−1)/(y+z))]_(y=0) ^2 +(1/2)[(1/((y+z+3)))]_(y=0) ^2 =−(1/2){(1/(2+z))−(1/z)} +(1/2){(1/(5+z))−(1/(3+z))} ⇒ ∫∫∫(...)dxdydz =−(1/2)∫_0 ^3 ((1/(z+2))−(1/z))dz+(1/2)∫_0 ^3 ((1/(5+z))−(1/(3+z)))dz =−(1/2)[ln∣((z+2)/z)∣]_0 ^3 +(1/2)[ln∣((z+5)/(z+3))∣]_0 ^3 =∞ it seems that this integral is divergent...!](https://www.tinkutara.com/question/Q75067.png)
$${x}+{z}=\mathrm{3}\:\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{3}\:{and}\:\mathrm{0}\leqslant{z}\leqslant\mathrm{3}\:\:\:{we}\:{have}\:\:\mathrm{0}\leqslant{y}\leqslant\mathrm{2}\:\Rightarrow \\ $$$$\int\int\int\:\:\frac{{dxdydz}}{\left({x}+{y}+{z}\right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{3}} \left(\:\int_{{o}} ^{\mathrm{2}} \:\:\left(\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\frac{{dx}}{\left({x}+{y}+{z}\right)^{\mathrm{3}} }\right){dy}\right){dz}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{{dx}}{\left({x}+{y}+{z}\right)^{\mathrm{3}} }\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+{z}\right)^{−\mathrm{2}} \right]_{{x}=\mathrm{0}} ^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\mathrm{3}+{y}+{z}\right)^{−\mathrm{2}} −\left({y}+{z}\right)^{−\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({y}+{z}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({y}+{z}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \left(\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\frac{{dx}}{\left({x}+{y}+{z}\right)^{\mathrm{3}} }\right){dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{dy}}{\left({y}+{z}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{dy}}{\left({y}+{z}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−\mathrm{1}}{{y}+{z}}\right]_{{y}=\mathrm{0}} ^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\left({y}+{z}+\mathrm{3}\right)}\right]_{{y}=\mathrm{0}} ^{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}+{z}}−\frac{\mathrm{1}}{{z}}\right\}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{5}+{z}}−\frac{\mathrm{1}}{\mathrm{3}+{z}}\right\}\:\Rightarrow \\ $$$$\int\int\int\left(…\right){dxdydz}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{3}} \left(\frac{\mathrm{1}}{{z}+\mathrm{2}}−\frac{\mathrm{1}}{{z}}\right){dz}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{5}+{z}}−\frac{\mathrm{1}}{\mathrm{3}+{z}}\right){dz} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{z}+\mathrm{2}}{{z}}\mid\right]_{\mathrm{0}} ^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{z}+\mathrm{5}}{{z}+\mathrm{3}}\mid\overset{\mathrm{3}} {\right]}_{\mathrm{0}} \\ $$$$=\infty\:\:\:{it}\:{seems}\:{that}\:{this}\:{integral}\:\:{is}\:{divergent}…! \\ $$
Answered by MJS last updated on 06/Dec/19
$$\int\frac{{dz}}{\left({x}+{y}+{z}\right)^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{2}\left({x}+{y}+{z}\right)^{\mathrm{2}} }+{C}_{\mathrm{1}} \\ $$$$\int\left(−\frac{\mathrm{1}}{\mathrm{2}\left({x}+{y}+{z}\right)^{\mathrm{2}} }+{C}_{\mathrm{1}} \right){dy}=\frac{\mathrm{1}}{\mathrm{2}\left({x}+{y}+{z}\right)}+{C}_{\mathrm{1}} {y}+{C}_{\mathrm{2}} \\ $$$$\int\left(\frac{\mathrm{1}}{\mathrm{2}\left({x}+{y}+{z}\right)}+{C}_{\mathrm{1}} {y}+{C}_{\mathrm{2}} \right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{x}+{y}+{z}\mid\:+{C}_{\mathrm{1}} {xy}+{C}_{\mathrm{2}} {x}+{C}_{\mathrm{3}} \\ $$