Chapter 8 - Further Techniques and Applications of Integration - 8.4 Improper Integrals - 8.4 Exercises - Page 452: 27

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\[\begin{align} & \int_{-\infty }^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx \\ & \int_{-\infty }^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx=\int_{-\infty }^{0}{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx+\int_{0}^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx \\ & \text{By the definition of improper integrals} \\ & \text{ }\int_{-\infty }^{0}{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx \\ & =-\frac{1}{4}\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{\left( -4{{x}^{3}} \right){{e}^{-{{x}^{4}}}}}dx \\ & =-\frac{1}{4}\underset{a\to -\infty }{\mathop{\lim }}\,\left[ {{e}^{-{{x}^{4}}}} \right]_{a}^{0} \\ & =-\frac{1}{4}\underset{a\to -\infty }{\mathop{\lim }}\,\left[ {{e}^{0}}-{{e}^{-{{a}^{4}}}} \right] \\ & \text{Evaluate when }a\to -\infty \\ & =-\frac{1}{4}\left[ 1-{{e}^{-\infty }} \right] \\ & =-\frac{1}{4} \\ & \text{ }\int_{0}^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx \\ & =-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\left( -4{{x}^{3}} \right){{e}^{-{{x}^{4}}}}}dx \\ & =-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-{{x}^{4}}}} \right]_{0}^{b} \\ & =-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-{{b}^{4}}}}-{{e}^{0}} \right] \\ & \text{Evaluate when }b\to \infty \\ & =-\frac{1}{4}\left[ {{e}^{-\infty }}-1 \right] \\ & =\frac{1}{4} \\ & \int_{-\infty }^{0}{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx+\int_{0}^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx=-\frac{1}{4}+\frac{1}{4} \\ & =0 \\ & \text{Therefore,} \\ & \int_{-\infty }^{\infty }{{{x}^{3}}{{e}^{-{{x}^{4}}}}}dx=0 \\ \end{align}\]