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Question Number 75027 by chess1 last updated on 06/Dec/19

Commented by mathmax by abdo last updated on 06/Dec/19
$x+z=3 ⇒0≤x≤3 and 0≤z≤3 we have 0≤y≤2 ⇒ ∫∫∫ ((dxdydz)/((x+y+z)^3 )) =∫_0 ^3 ( ∫_o ^2 (∫_0 ^3 (dx/((x+y+z)^3 )))dy)dz we have ∫_0 ^3 (dx/((x+y+z)^3 )) =[−(1/2)(x+y+z)^(−2) ]_(x=0) ^3 =−(1/2){(3+y+z)^(−2) −(y+z)^(−2) } =(1/(2(y+z)^2 ))−(1/((y+z+3)^2 )) ⇒ ∫_0 ^2 (∫_0 ^3 (dx/((x+y+z)^3 )))dy =(1/2) ∫_0 ^2 (dy/((y+z)^2 ))−(1/2)∫_0 ^2 (dy/((y+z+3)^2 )) =(1/2)[((−1)/(y+z))]_(y=0) ^2 +(1/2)[(1/((y+z+3)))]_(y=0) ^2 =−(1/2){(1/(2+z))−(1/z)} +(1/2){(1/(5+z))−(1/(3+z))} ⇒ ∫∫∫(...)dxdydz =−(1/2)∫_0 ^3 ((1/(z+2))−(1/z))dz+(1/2)∫_0 ^3 ((1/(5+z))−(1/(3+z)))dz =−(1/2)[ln∣((z+2)/z)∣]_0 ^3 +(1/2)[ln∣((z+5)/(z+3))∣]_0 ^3 =∞ it seems that this integral is divergent...!$
$x + z = 3 \Rightarrow 0 ⩽ x ⩽ 3 a n d 0 ⩽ z ⩽ 3 w e h a v e 0 ⩽ y ⩽ 2 \Rightarrow$ $\int \int \int \frac{d x d y d z}{{(x + y + z)}^{3}} = \int_{0}^{3} (\int_{o}^{2} (\int_{0}^{3} \frac{d x}{{(x + y + z)}^{3}}) d y) d z w e h a v e$ $\int_{0}^{3} \frac{d x}{{(x + y + z)}^{3}} = {[- \frac{1}{2} {(x + y + z)}^{- 2}]}_{x = 0}^{3} = - \frac{1}{2} {{(3 + y + z)}^{- 2} - {(y + z)}^{- 2}}$ $= \frac{1}{2 {(y + z)}^{2}} - \frac{1}{{(y + z + 3)}^{2}} \Rightarrow$ $\int_{0}^{2} (\int_{0}^{3} \frac{d x}{{(x + y + z)}^{3}}) d y = \frac{1}{2} \int_{0}^{2} \frac{d y}{{(y + z)}^{2}} - \frac{1}{2} \int_{0}^{2} \frac{d y}{{(y + z + 3)}^{2}}$ $= \frac{1}{2} {[\frac{- 1}{y + z}]}_{y = 0}^{2} + \frac{1}{2} {[\frac{1}{(y + z + 3)}]}_{y = 0}^{2}$ $= - \frac{1}{2} {\frac{1}{2 + z} - \frac{1}{z}} + \frac{1}{2} {\frac{1}{5 + z} - \frac{1}{3 + z}} \Rightarrow$ $\int \int \int (. . .) d x d y d z = - \frac{1}{2} \int_{0}^{3} (\frac{1}{z + 2} - \frac{1}{z}) d z + \frac{1}{2} \int_{0}^{3} (\frac{1}{5 + z} - \frac{1}{3 + z}) d z$ $Missing \left or extra \right$ $= \infty i t s e e m s t h a t t h i s i n t e g r a l i s d i v e r g e n t . . .!$
	Answered by MJS last updated on 06/Dec/19

	$\int \frac{d z}{{(x + y + z)}^{3}} = - \frac{1}{2 {(x + y + z)}^{2}} + C_{1}$ $\int (- \frac{1}{2 {(x + y + z)}^{2}} + C_{1}) d y = \frac{1}{2 (x + y + z)} + C_{1} y + C_{2}$ $\int (\frac{1}{2 (x + y + z)} + C_{1} y + C_{2}) d x = \frac{1}{2} \ln ∣ x + y + z ∣ + C_{1} x y + C_{2} x + C_{3}$
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