Question Number 220891 by Nicholas666 last updated on 20/May/25
![∫∫∫_([0,∞]^( 3) ) (e^(−(x + y + z )) /(1 + xyz)) dxdydz](Q220891.png)
$$ \\ $$$$\:\:\:\:\:\int\int\int_{\left[\mathrm{0},\infty\right]^{\:\mathrm{3}} } \:\frac{{e}^{−\left({x}\:+\:{y}\:+\:{z}\:\right)} }{\mathrm{1}\:+\:{xyz}}\:{dxdydz} \\ $$$$ \\ $$
Answered by breniam last updated on 20/May/25
![=∫_0 ^1 ∫_0 ^∞ e^(−z) ∫_0 ^∞ e^(−y) ∫_0 ^∞ w^(xyz) e^(−x) dxdydzdw= =∫_0 ^1 ∫_0 ^∞ e^(−z) ∫_0 ^∞ e^(−y) ∫_0 ^∞ ((w^(yz) /e))^x dxdydzdw= −∫_0 ^1 ∫_0 ^∞ e^(−z) ∫_0 ^∞ e^(−y) ln((w^(yz) /e))dydzdw= ∫_0 ^1 ∫_0 ^∞ e^(−z) [∫_0 ^∞ e^(−y) dy−zln(w)∫_0 ^∞ ye^(−y) dy]dzdw= ∫_0 ^1 ∫_0 ^∞ e^(−z) [1−zln(w)]dzdw=∫_0 ^1 [∫_0 ^∞ e^(−z) dz−ln(w)∫_0 ^∞ ze^(−z) dz]dw= ∫_0 ^1 [1−ln(w)]dw=1−∫_0 ^1 w′ln(w)dw=1+∫_0 ^1 dw=2](Q220926.png)
$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}} \underset{\mathrm{0}} {\overset{\infty} {\int}}{w}^{{xyz}} {e}^{−{x}} \mathrm{d}{x}\mathrm{d}{y}\mathrm{d}{z}\mathrm{d}{w}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}} \underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{{w}^{{yz}} }{{e}}\right)^{{x}} \mathrm{d}{x}\mathrm{d}{y}\mathrm{d}{z}\mathrm{d}{w}= \\ $$$$−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}} \mathrm{ln}\left(\frac{{w}^{{yz}} }{{e}}\right)\mathrm{d}{y}\mathrm{d}{z}\mathrm{d}{w}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \left[\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}} \mathrm{d}{y}−{z}\mathrm{ln}\left({w}\right)\underset{\mathrm{0}} {\overset{\infty} {\int}}{ye}^{−{y}} \mathrm{d}{y}\right]\mathrm{d}{z}\mathrm{d}{w}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \left[\mathrm{1}−{z}\mathrm{ln}\left({w}\right)\right]\mathrm{d}{z}\mathrm{d}{w}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \mathrm{d}{z}−\mathrm{ln}\left({w}\right)\underset{\mathrm{0}} {\overset{\infty} {\int}}{ze}^{−{z}} \mathrm{d}{z}\right]\mathrm{d}{w}= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\mathrm{1}−\mathrm{ln}\left({w}\right)\right]\mathrm{d}{w}=\mathrm{1}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{w}'\mathrm{ln}\left({w}\right)\mathrm{d}{w}=\mathrm{1}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{d}{w}=\mathrm{2} \\ $$