Question Number 219354 by SdC355 last updated on 23/Apr/25
$$\int_{−\infty} ^{\:+\infty} \:\:{ze}^{−{z}^{\mathrm{3}} } \:\mathrm{d}{z}=?? \\ $$
Answered by breniam last updated on 24/Apr/25
$$\underset{−\infty} {\overset{\mathrm{0}} {\int}}{ze}^{−{z}^{\mathrm{3}} } \mathrm{d}{z}=\left\{\overset{−} {{z}}=−{z}\right\}=−\underset{\mathrm{0}} {\overset{\infty} {\int}}{ze}^{{z}^{\mathrm{3}} } \mathrm{d}{z}\leqslant−\underset{\mathrm{0}} {\overset{\infty} {\int}}{ze}^{{z}^{\mathrm{2}} } \mathrm{d}{z}=\left\{\overset{−} {{z}}={z}^{\mathrm{2}} \right\}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{{z}} \mathrm{d}{z}=−\infty \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{ze}^{−{z}^{\mathrm{3}} } \mathrm{d}{z}\leqslant\underset{\mathrm{0}} {\overset{\infty} {\int}}{ze}^{−{z}^{\mathrm{2}} } \mathrm{d}{z}=\left\{\overset{−} {{z}}={z}^{\mathrm{2}} \right\}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}} \mathrm{d}{z}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Integral}\:\mathrm{diverges} \\ $$